The function y = 1.5(1.4)^x can model exponential growth or decay.
The number 1.4 represents the growth or decay factor, and the number 1.5 represents the initial quantity.
To solve the equation 6.2 = 1.5(1.4)^x, we find an exact solution and a decimal estimate, which represents the time when the quantity reaches 6.2 in the given situation.
1a. The function y = f(x) = 1.5(1.4)^x can model the situation of exponential growth or decay. For example, it could represent the population of bacteria in a culture over time, where x is the time in hours, y is the number of bacteria, and 1.4 represents the growth factor of 40% per unit of time.
1b. In this situation, the number 1.4 represents the growth factor or decay factor per unit of time. It indicates how much the quantity is increasing or decreasing at each step of the time interval.
1c. The number 1.5 in the equation represents the initial quantity or starting value of the situation being modeled. It is the value of y when x = 0 or the initial condition of the scenario.
1d. To solve the equation 6.2 = 1.5(1.4)^x:
Divide both sides of the equation by 1.5:
(1.4)^x = 6.2/1.5
Take the logarithm (base 1.4) of both sides:
x = log(6.2/1.5) / log(1.4)
This is the exact solution. To find a decimal estimate, evaluate the expression using a calculator:
x ≈ 3.663
In this situation, the decimal estimate of x = 3.663 represents the time at which the quantity reaches the value of 6.2 based on the given exponential growth or decay model.
1e. To check the solution from part (1d) using a table or graph:
Table: Generate a table of values for the function y = 1.5(1.4)^x for various x values. Evaluate the function for x = 3.663 and see if it gives a value close to 6.2.
Graph: Plot the function y = 1.5(1.4)^x on a graphing calculator or software. Locate the point where the graph intersects the y = 6.2 line and check if it aligns with the estimated x value.
Both methods will allow you to visually and numerically verify if the x value obtained from solving the equation matches the desired y value of 6.2.
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As discussed in class (and in Hartmann Chapter 4) the exchange of heat and momentum between the atmosphere and the earth surface can be computed using the aerodynamic formula: T = = PcU? CPCU.(T. -T.) (5) SH = LE = LPCU.(9-9-) where UT, and q, are the wind speed, temperature and specific humidity respectively at a reference height (usually 10 m) above the earth surface. T. and q, are respectively the temperature and specific humidity at the surface, co is the aerodynamic exchange coefficient, and L is the latent heat of vaporization of water. (a) Compute the wind stress T, sensible heat flux (SH) and latent heat of evaporation (LE) from the land surface when T. = 30°C, T, = 28°C, 45 = 1.6 x 10-29, = 1.5 x 10-- and U. = 5 ms! Let's assume that the atmospheric boundary layer is unstable (as during the daytime when the land surface is warmer than the air above) so that co = 4 x 10-2. In each case state whether the direction of the flur is toward or away from the surface, and provide the appropriate units for t, SH and LE. (10 points) = = (6) Now repeat the calculations in (a) but over the ocean for T, = 28°C, T, = 30°C, 9. 1.6 x 10-29, = 1.5 x 10-2 and U, = 5 m st. In this case the atmospheric boundary layer is generally stable because the surface is cooler than the air above so co = 1 x 10- In each case state whether the direction of the flux is toward or away from the surface, and provide the appropriate units for T, SH and LE. (5 points) (C) State what kind of turbulence to you would expect to be dominant in the atmospheric boundary layer in each case. (4 points). If the value of a constant or parameter is not given, you will need to look it up in the textbook or online.
(a) T = 24 N/m² (toward surface), SH = 120 W/m² (away from surface), LE = 800 W/m² (away from surface).
(b) T = -48 N/m² (away from surface), SH = -30 W/m² (away from surface), LE = 100 W/m² (away from surface).
(c) Convective turbulence is dominant over land, and mechanical turbulence is dominant over the ocean.
To compute the wind stress (T), sensible heat flux (SH), and latent heat of evaporation (LE) from the land surface, we'll use the given equations and provided values:
(a) Land Surface Calculations:
T. = 30°C, T = 28°C, 45 = [tex]1.6 * 10^{-2}[/tex], Ω = 1.5 x [tex]10^{-2}[/tex], U = 5 m/s, and co = 4 x [tex]10^{-2}[/tex].
Using Equation (5) for T:
T = ρcU²(T. - T) = (1.2 kg/m³)(4 x 10^-2)(5 m/s)²(30°C - 28°C) = 24 N/m² (toward surface)
Using Equation (5) for SH:
SH = LPCU(T. - T) = ([tex]1.5 x 10^3[/tex]J/kg)([tex]4 * 10^{-2}[/tex])(30°C - 28°C) = 120 W/m² (away from surface)
Using Equation (5) for LE:
LE = LPCU(q*. - q) = (2.5 x [tex]10^6[/tex] J/kg)([tex]4 x 10^{-2}[/tex])(1.6 x [tex]10^{-2}[/tex] - 1.5 x [tex]10^{-2}[/tex]) = 800 W/m² (away from surface)
(b) Ocean Surface Calculations:
T. = 28°C, T = 30°C, 45 = 1.6 x [tex]10^{-2}[/tex], Ω = 1.5 x [tex]10^{-2}[/tex], U = 5 m/s, and co = 1 x [tex]10^{-2}[/tex].
Using Equation (5) for T:
T = ρcU²(T. - T) = (1.2 kg/m³)(1 x [tex]10^{-2}[/tex])(5 m/s)²(28°C - 30°C) = -48 N/m² (away from surface)
Using Equation (5) for SH:
SH = LPCU(T. - T) = (1.5 x [tex]10^3[/tex] J/kg)(1 x [tex]10^{-2}[/tex])(28°C - 30°C) = -30 W/m² (away from surface)
Using Equation (5) for LE:
LE = LPCU(q*. - q) = (2.5 x [tex]10^6[/tex] J/kg)(1 x [tex]10^{-2}[/tex])(1.6 x [tex]10^{-2}[/tex] - 1.5 x [tex]10^{-2}[/tex]) = 100 W/m² (away from surface)
(c) Dominant Turbulence:
In the unstable atmospheric boundary layer over land, convective turbulence is expected to be dominant due to the warmer surface heating the air above.
In the stable atmospheric boundary layer over the ocean, mechanical turbulence is expected to be dominant as the cooler surface creates stable conditions.
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6. How do you find the slope of a line?
A Divide the difference of the y values by the differences of the x values.
B Divide the difference of the x values by the differences of the y values.
C Divide the rise of the line by the run of the line.
D Both A and C.
Answer:
D
Step-by-step explanation:
A group of 12 students participated in a dance competition. Their scores are below:
Score (points)
1
2
3
4
5
Number of Students
1
2
4
3
2
Would a dot plot or a histogram best represent the data presented here? Why? (3 points)
Group of answer choices
Histogram, because a large number of scores are reported as ranges
Histogram, because a small number of scores are reported individually
Dot plot, because a large number of scores are reported as ranges
Answer:
put a picture please so i can understand better
Answer:
The answer is A
Step-by-step explanation:
i have done this and got 100 :0
A. 3
B. 10
C. 20
D. 30
Answer:
D. 30Explanation:
292/100×9.9 = 2.92×9.9 = 28.908
30 is closest to 28.908
A silo is a composite of a cylindrical tower with a cone for a roof. What is the volume of the silo if the radius of the base is 40 feet, the height of the roof is 10 feet, and the height of the entire silo is 75 feet?
a. 393,746.3 ft
b. 343,480.8 ft
c. 326,725.6 ft
d. 376,991.1 ft
Answer:
The answer is b. 343,480.8ft
Step-by-step explanation:
I thought it was c at first because I forgot to add the volume of the cone.
The equation for the volume of a cylinder is V=π×r²×h
The solution would look like V=π40²×65=326725.8
The equation for the volume of a cone is V=1/3πr²×h
The solution would look like V=1/3π×40²×10=16755
Adding the two volumes would equal 326725.8±16755= 343480.8
A principal was buying T-shirts for his school's chess club and found that the total cost
in dollars could be found by the function f(x) = 9x + 19, where x is the number of
members in the club. If there are at least 8 members on the team but not more than
11, then which of the following statements describes the function?
A. The value of x must be a whole number between 8 and 11 and the value of f(x) must
be a whole number between 72 and 99.
B. The value of x must be a whole number between 8 and 11 and the value of f(x) must
be a whole number between 91 and 118.
C. The values of x and f(x) must both be whole numbers between 91 and 118.
D. The values of x and f(x) must both be whole numbers between 9 and 19.
Triangle ABC below has a right angle C and a height CD. (A height in a triangle connects its vertex and the opposite side, and is perpendicular to that side.)
If AC is 3 units long and BC is 4 units long, what is the length of AD?
can you show the picture please?
If the unit rate is constant, what are the total sales for 12 pounds of asparagus?
Evaluate the following integrals. a. if R is the rectangle R = [0,3] x [0,1/2). = [ xsinº y dA R b. if D is the region bounded by the y-axis, yox, and yu4 If y²er dA.
a. The value of the integral ∬R xsin(θ) dA over the rectangle R = [0,3] x [0,1/2) is (9/4) sin(θ).
b. The value of the integral ∬D y^2 dA over the region bounded by the y-axis, y = 0, and y = 4 is 64.
a. To evaluate the integral ∬R xsin(θ) dA over the rectangle R = [0,3] x [0,1/2), we can use Cartesian coordinates. The integral becomes:
∬R xsin(θ) dA = ∫[0,3] ∫[0,1/2] x sin(θ) dy dx
Integrating with respect to y first:
∫[0,3] ∫[0,1/2] x sin(θ) dy dx = ∫[0,3] [x sin(θ) y] [0,1/2] dx
= ∫[0,3] (x sin(θ) (1/2 - 0)) dx
= ∫[0,3] (x sin(θ)/2) dx
= (sin(θ)/2) ∫[0,3] x dx
= (sin(θ)/2) [x^2/2] [0,3]
= (sin(θ)/2) (9/2)
= (9/4) sin(θ)
b. To evaluate the integral ∬D y^2 dA over the region bounded by the y-axis, y = 0, and y = 4, we can use Cartesian coordinates. The integral becomes:
∬D y^2 dA = ∫[0,4] ∫[0,y] y^2 dx dy
Integrating with respect to x first:
∫[0,4] ∫[0,y] y^2 dx dy = ∫[0,4] [y^2 x] [0,y] dy
= ∫[0,4] (y^3 - 0) dy
= ∫[0,4] y^3 dy
= [y^4/4] [0,4]
= 4^4/4 - 0
= 64
Therefore, the value of the integral ∬D y^2 dA is 64.
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match the answer to the equation
1: no solution
2: infinitely many solutions
3: one solution
Michelle counted 40 green cars and 20 silver cars in the parking lot. If
the number of green cars stays the same, how many more silver cars
would need to be added so the ratio of green cars to silver cars is 1 to 3?
Answer:
100 more silver cars
Step-by-step explanation:
Right now the ratio of silver cars to green cars is 20:40 or 1:2, but if you add a hundred more silvers cars the ratio would 120:40 or 3:1.
QUESTION 4
Factorise fully:
4.1.1
12x² – x - 6
Answer:
y = (4x - 3)(3x + 2)
Step-by-step explanation:
Factorization is the process of taking like terms out of number such that one can rewrite an expression by its factors, or as the products of serval smaller expressions.
The first step to factoring an equation is to rewrite the linear term as the sum or difference of two other linear terms. One must do this in such as way that it shares a common factor with one of the other terms in the expression.
[tex]12x^2 -x - 6[/tex]
[tex]12x^2 -9x + 8x - 6[/tex]
Take out the common factor,
[tex]3x(4x-3)+2(4x-3)[/tex]
Factor,
[tex](3x+2)(4x-3)[/tex]
help me pls & thank you:)
Answer:the answer is 53 degrees
160 overall degrees 3 numerals divide 160 by 3 and it's 53.33 but it's degrees so round to the nearest tenth so it stays as 53 degrees
Step-by-step explanation:
Norma builds a 1/8 scale Model of her own house.Her living room measures 12 feet by 28 feet. What are the dimensions of the models living room?
Answer:
1.5 feets by 3.5 feets
Step-by-step explanation:
Scale model = 1/8
Dimension of living room = 12 feets by 28 feets
The dimension of the model using the scale model given :
Scale model * dimension
1/8 * (12 feets by 28feets)
= 1.5 feets by 3.5 feets
For a new study conducted by a fitness magazine, 260 females were randomly selected. For each, the mean daily calorie consumption was calculated for a September-February period. A second sample of 300 females was chosen independently of the first. For each of them, the mean daily calorie consumption was calculated for a March-August period. During the September-February period, participants consumed a mean of 2387.1 calories daily with a standard deviation of 210. During the March-August period, participants consumed a mean of 2412.9 calories daily with a standard deviation of 267.5. The population standard deviations of daily calories consumed for females in the two periods can be estimated using the sample standard deviations, as the samples that were used to compute them were quite large. Construct a 90% confidence interval for μ1−μ2, the difference between the mean daily calorie consumption of μ1 females in September-February and the mean daily calorie consumption of μ2 females in March-August.
Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)
What is the lower limit of the 90% confidence interval?
What is the upper limit of the 90% confidence interval?
the lower limit of the 90% confidence interval is approximately -65.25, and the upper limit is approximately 13.65.
To construct a 90% confidence interval for the difference between the mean daily calorie consumption of females in September-February (μ₁) and the mean daily calorie consumption of females in March-August (μ₂), we can use the formula:
CI = ([tex]\bar{X_1}[/tex] - [tex]\bar{X_2}[/tex]) ± Z * √((s₁² / n₁) + (s₂² / n₂))
Where:
[tex]\bar{X_1}[/tex] and [tex]\bar{X_2}[/tex] are the sample means of calorie consumption for the two periods.
s₁ and s₂ are the sample standard deviations of calorie consumption for the two periods.
n₁ and n₂ are the sample sizes for the two periods.
Z is the z-score corresponding to the desired confidence level.
Given data:
[tex]\bar{X_1}[/tex] = 2387.1 (mean daily calorie consumption for September-February)
[tex]\bar{X_2}[/tex] = 2412.9 (mean daily calorie consumption for March-August)
s₁ = 210 (standard deviation for September-February)
s₂ = 267.5 (standard deviation for March-August)
n₁ = 260 (sample size for September-February)
n₂ = 300 (sample size for March-August)
Confidence level = 90%
First, we need to find the z-score corresponding to a 90% confidence level. The z-score can be found using a z-table or a calculator. For a 90% confidence level, the z-score is approximately 1.645.
Now, we can substitute the values into the formula to calculate the confidence interval:
CI = (2387.1 - 2412.9) ± 1.645 * √((210² / 260) + (267.5² / 300))
Calculating the values inside the square root:
√((210² / 260) + (267.5² / 300)) ≈ √(342.461538 + 238.083333) ≈ √(580.544872) ≈ 24.107
Substituting the values into the formula:
CI = -25.8 ± 1.645 * 24.107
Calculating the limits of the confidence interval:
Lower limit = -25.8 - 1.645 * 24.107 ≈ -65.246
Upper limit = -25.8 + 1.645 * 24.107 ≈ 13.646
Rounding the values to two decimal places:
Lower limit ≈ -65.25
Upper limit ≈ 13.65
Therefore, the lower limit of the 90% confidence interval is approximately -65.25, and the upper limit is approximately 13.65.
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(10-4)²(12+10/2)=_________
Answer:
should be 612
Step-by-step explanation:
Answer:
(10-4)²(12+10/2)=
(6)²(12+5)=
(36) (17) = 612
•°|NO LINKS PLEASE, REAL ANSWERS ONLY!|°•
The dimensions of two pyramids formed of sand are shown. How much more sand is in the pyramid with the greater volume?
There are __ more cubic inches of sand in the pyramid with the greater volume.
Answer:
V=75 in 3 9 in. 17 in. There are 1 more cubic inches of sand in the pyramid with the greater volume.
Step-by-step explanation:
mr. and mrs. simpson went to two movies. the first movie lasted 2 ⅓ hours and the second one lasted 1 ⅘ hours. how much longer was the first than the second movie?
The first movie was ⅔ hours longer than the second one.
The difference in length between the two movies, we need to subtract the length of the second movie from the first.
we need to convert the mixed numbers to improper fractions to make the subtraction easier.
2 ⅓ hours can be written as 7/3 hours.
1 ⅘ hours can be written as 8/5 hours.
Now, we can subtract them
7/3 - 8/5
we need a common denominator. The least common multiple of 3 and 5 is 15.
(7/3) * (5/5) - (8/5) * (3/3) = 35/15 - 24/15 = 11/15
Therefore, the first movie was 11/15 hours (or approximately 44 minutes) longer than the second one
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Genoude towing test at the 0.10 vel tance by temperative contested to Put Assure at the comptes were obland de condendy ang simple random samping Test whether PPSample data are 30, 256, 38, 31 (a) Determine the rulanternative hypotheses. Choose the correct answer below OAH, versus H P = 0 OCH PD, versus HDD) OBH PP, VOUS HP) OD HIPP versus HDP (b) The fest statistics found to two decimal places as needed (c) The P values Round to three decimal places as needed) Test meil hypothesis. Choose the correct condusion below OA Roth null hypothesis because there is not suit evidence to conclude that, (b) The test statistic Zo is (Round to two decimal places as needed.) (c) The P-value is (Round to three decimal places as needed.) Test the null hypothesis. Choose the correct conclusion below. O A. Reject the null hypothesis because there is not sufficient evidence to conclude that p, p2. OC. Do not reject the null hypothesis because there is not sufficient evidence to conclude that p, #P2. OD. Reject the null hypothesis because there is sufficient evidence to conclude that p, p2.
The given problem involves hypothesis testing in statistics. The correct conclusion will depend on whether the null hypothesis is rejected or not.
(a) The null and alternative hypotheses can be determined as follows: OAH (One-sample test for proportion): P = 0 (H0) versus H P ≠ 0 (HA).
(b) The test statistic, denoted as Zo, needs to be computed using the given sample data.
(c) To determine the P-value, the calculated test statistic is compared to the appropriate distribution (e.g., standard normal distribution) based on the chosen significance level.
(d) Based on the P-value and the predetermined significance level, the null hypothesis is either rejected or not rejected. If the P-value is less than the significance level, the null hypothesis is rejected. Otherwise, the null hypothesis is not rejected.
The conclusion will depend on whether the null hypothesis is rejected or not and should be stated accordingly.
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Prove that in n , a single point {} is a closed
sets
please write down your answer in detail.
To prove that a single point {a} is a closed set in Rⁿ, we need to show that its complement, denoted as Rⁿ \ {a}, is open.
Let's consider an arbitrary point x in the complement Rⁿ \ {a}. Since x is not equal to a, there exists a positive radius r such that the open ball B(x, r) centered at x with radius r does not contain a.
Now, let's show that B(x, r) is entirely contained within Rⁿ\ {a}. We need to demonstrate that for any point y in B(x, r), y is also in Rⁿ \ {a}.
If y is equal to x, then y is not equal to a since a is excluded from Rⁿ \ {a}. Therefore, y is in Rⁿ\ {a}.
If y is not equal to x, then we can consider the distance between y and a. Since y is in B(x, r), we have:
d(y, x) < r
However, since a is not in B(x, r), we have:
d(a, x) ≥ r
Now, let's consider the distance between y and a:
d(y, a) ≤ d(y, x) + d(x, a) < r + (d(a, x) - r) = d(a, x)
Since d(y, a) is strictly less than d(a, x), it follows that y is not equal to a. Therefore, y is in Rⁿ \ {a}.
This shows that for every point x in Rⁿ \ {a}, there exists an open ball B(x, r) that is entirely contained within Rⁿ \ {a}. Hence, Rⁿ\ {a} is open.
By the definition of a closed set, if the complement of a set is open, then the set itself is closed. Therefore, a single point {a} is a closed set in Rⁿ.
The question should be:
Prove that in Rⁿ , a single point {a} is a closed sets.please write your answer in detail
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use the law of cosines to find the missing angle. find m∠a to the nearest tenth of a degree
The law of cosines can be used to find the missing angle when the lengths of three sides of a triangle are known. In this case, we want to find the measure of angle a, so we'll use the formula: cos(a) = (b² + c² - a²) / (2bc)where a is the side opposite angle a, b is the side opposite angle b, and c is the side opposite angle c.
To find angle a, we need to rearrange the formula: cos(a) = (b² + c² - a²) / (2bc)cos(a) = (7² + 13² - 10²) / (2 * 7 * 13)cos(a) = 0.81923077a = cos⁻¹(0.81923077)a ≈ 34.2°Therefore, m∠a is approximately 34.2 degrees.
When either the lengths of the two sides and the measure of the included angle (SAS) or the lengths of the three sides (SSS) are known, the Law of Cosines is used to determine the remaining parts of an oblique (non-right) triangle.
The square of a side of an oblique triangle is equal to the sum of the squares of the other two sides minus twice the product of the two sides if A, B, and C are the measures of the angles of the triangle and a, b, and c are the lengths of the sides opposite the corresponding angles.
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If the figure shown on the grid below is dilated by a scale factor of 2/3 with the center of dilation at (-4,4), what is the coordinate of point M after the dilation?
Step-by-step explanation:
To determine the coordinate of point M after the dilation, we need to apply the scale factor and center of dilation to the original coordinates.
Given:
Scale factor = 2/3
Center of dilation = (-4, 4)
Let's assume the coordinates of point M in the original figure are (x, y). To find the new coordinates after dilation, we can use the following formula:
New x-coordinate = Center of dilation x-coordinate + (Original x-coordinate - Center of dilation x-coordinate) * Scale factor
New y-coordinate = Center of dilation y-coordinate + (Original y-coordinate - Center of dilation y-coordinate) * Scale factor
Substituting the given values, we have:
New x-coordinate = (-4) + (x - (-4)) * (2/3)
New y-coordinate = 4 + (y - 4) * (2/3)
Since we are specifically looking for the coordinate of point M after dilation, we can substitute M's original coordinates into the formulas. Let's assume the original coordinates of point M are (xM, yM):
New x-coordinate = (-4) + (xM - (-4)) * (2/3)
New y-coordinate = 4 + (yM - 4) * (2/3)
Now we have the coordinates of point M after the dilation.
Please provide the values of xM and yM to calculate the specific coordinate of point M after the dilation.
Answer:
What does it mean to dilate by a scale factor of 3?
The key thing is that the dilation value affects the distance between two points. As in the first example (dilation by a factor of 3), A is originally 1 unit down from P and 2 units to the left of P. 1*3 = 3, so A' (the dilated point) should be 3 unit
Step-by-step explanation:
Give the similarity ratio for the following figures (small to large).
Answer: I think the answer would be 1/6
Step-by-step explanation:
If you notice, all of the numbers on the left model are 1/6 less than the one on the left. I don't know what type of answer they are looking for but the one on the left is 1.6 smaller than the one on the right:)
Evaluate (-2xy^3)^3 when x = 4 and y = -1
Step-by-step explanation:
(-2xy^3)^3
(-2(4)(-1)^3)^3
(-2(4)(-1))^3
(8)^3
512
Number of participants = 274 Number of groups = 5 SSD = 1 SSW = 192 What is the mean squares within?
The mean squares within is 0.7130044843.
The formula to calculate the mean squares within is given below:
Mean squares within (MSW) = SSW / (n - k), where n = Total number of observations, k = Total number of groups.
Here, Number of participants = 274, Number of groups = 5, SSD = 1, SSW = 192.
Now, the value of n and k can be calculated using the formula given below:
n = Number of participants = 274, k = Number of groups = 5.
Let's substitute the values in the formula of mean squares within:
Mean squares within (MSW) = SSW / (n - k)
MSW = 192 / (274 - 5)
MSW = 192 / 269
MSW = 0.7130044843
Hence, the mean squares within is 0.7130044843.
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PLEASE HELP ME OUT! QUICK POINTS FOR YOU!
All information needed can be found in the image below
Thank you in advance.
Answer:
3.14 units squared
Step-by-step explanation:
A = pi r^2
A = pi (1) ^2
A = pi = 3.14
(Radius = diameter / 2)
Find the surface area of the triangular prism. The base of the prism is an isosceles triangle. The answer is _ cm^2
Thanks in advance!
Answer:
3408cm²
Step-by-step explanation:
Please help me I will mark...BRAINLIEST
I need volume and surface area
Answer:
1. V=60m³ SA=94m²
2. V=32m³ SA=88m²
3. V=2000mm³ SA=1000mm²
4. SA=376.8cm² V=552.64cm³
5. SA=1130.4cm² V=1356.48mm³
Step-by-step explanation:
1. Volume=5*4*3=60m³
Surface Area=2*(3*4)+2*(5*3)+2*(4*5)=24+30+40=94m²
2. Volume=8*4*1=32cm³
Surface Area=2*(8*1)+2*(8*4)+2*(4*1)=16+64+8=88cm²
3. Volume=10*10*20=2000mm³
Surface Area=2*(10*10)+2*(20*10)+2*(20*10)=200+400+400=1000mm²
4. SA of the cylinder=(2πr*h)+(2*πr²)=(2*(3.14)*4*11)+(2*3.14*16)
=(276.32)+(100.48)=376.8cm²
Volume=hπr²=11*(3.14)*16=552.64cm³
5. SA of cylinder=(2πr*h)+(2*πr²)=(2*(3.14)*12*3)+(2*3.14*144)
=(904.32)+(226.08)=1130.4mm²
Volume=hπr²=3*(3.14)*144=1356.48mm³
6 and 186 its divison i dont know division
[tex]\sqrt{x}[/tex]
Answer:
31
Step-by-step explanation:
u divide 186 by 6. u divide by pretty much seeing how many times 6 can go into 186.
Answer:
31
Step-by-step explanation:
assuming your asking "what is 186 divided by 6?" ...right?
SQUARE ROOT OF 21316 WITH STEP BY STEP SOLUTION THANKS
Answer:
146
Step-by-step explanation:
[tex] = \sqrt{21316 } [/tex]
[tex] = \sqrt{4 \times 5329} [/tex]
[tex] = 2 \sqrt{5329} [/tex]
[tex] = 2 \sqrt{73 \times 73} [/tex]
[tex] = 2 \times 73[/tex]
[tex] = 146[/tex]