if a=4 b=6 and c=10 what is the volume of the three prisms above

Answer: x=4

Step-by-step explanation:

First find the smallest x which would be 5x and subtract that from the 9x. So now it should be 4x-4=12. Then, move 4 and add that to the twelve which should now be 4x=16. Then, divide it by four and you get x=4.

Answer:

x = 4

Step-by-step explanation:

We start from

9x - 4 = 5x + 12

We move the "5x" to the left side of the "=" sign by changing it's sign from 5x which can be seen as +5x to -5x

9x - 4 - 5x = 12

Now we move the "4" to the right side of the "=" sign by applying the same principle as above, so -4 turns to +4

9x - 5x = 12 + 4

We than solve the subtraction and addition

4x = 16

Then we divide both sides of the equation by 4 and we get

x = 4

The number of ways in which three digits can be written is 720 ways.

The number of ways in which 3 letters can be written is 15600.

From the information, the subject identification numbers in a certain scientific research project consist of two digits followed by four letters and then two more digits.

The ways in which three digits can be written will be:.

= 10P 3

= 10! / (10-3)!

= 10 × 9 × 8

= 720 ways

The number of ways in which 3 letters can be written will be:

= 26P3

= 26! / ( 26 - 3)!

= 26! / 23!

= 26 × 25 × 24

= 15600

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An estimate of the quotient is in improper fraction [tex]\frac{133}{60}[/tex]. An estimate of the quotient is in mixed fraction [tex]2 \frac{13}{60}$$[/tex]

[tex]6 \frac{1}{3} \div 2 \frac{6}{7}$$[/tex]

Convert mixed numbers to improper fractions: [tex]$\quad 6 \frac{1}{3}=\frac{19}{3}$[/tex]

[tex]=\frac{19}{3} \div 2 \frac{6}{7}$$[/tex]

Convert mixed numbers to improper fractions: [tex]$2 \frac{6}{7}=\frac{20}{7}[/tex]

[tex]$=\frac{19}{3} \div \frac{20}{7}$[/tex]

Apply the fraction rule: [tex]$\frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \times \frac{d}{c}$[/tex]

[tex]=\frac{19}{3} \times \frac{7}{20}$$[/tex]

Multiply fractions: [tex]$\frac{a}{b} \times \frac{c}{d}=\frac{a \times c}{b \times d}$[/tex]

[tex]=\frac{19 \times 7}{3 \times 20}$$[/tex]

Multiply fractions:[tex]\frac{a}{b} \times \frac{c}{d}=\frac{a \times c}{b \times d}$[/tex]

[tex]=\frac{19 \times 7}{3 \times 20}$$[/tex]

Multiply the numbers: [tex]$19 \times 7=133$[/tex]

[tex]=\frac{133}{3 \times 20}$$[/tex]

Multiply the numbers: [tex]$3 \times 20=60$[/tex]

[tex]=\frac{133}{60}$$[/tex]

Convert improper fractions to mixed numbers: [tex]$\frac{133}{60}=2 \frac{13}{60}$[/tex]

[tex]=2 \frac{13}{60}$$[/tex]

When it comes to adding Mixed or Improper fractions, we can have either the same denominators for both the fractions to be added or the denominators can differ too.

A mixed fraction is one that is represented by both its quotient and remainder. A mixed fraction is, for instance, 2 1/3, where 2 is the quotient and 1 is the remainder. An amalgam of a whole number and a legal fraction is a mixed fraction.

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Given:

(12, 8) and (X, -4 )

Required:

To calculate the value of x

Explanation:

Determine the parameter through distance formula

Required answer:

x=7 or x=17

Using transformations, more specifically a translation, the functions that could represent the one graphed are given as follows:

We first consider the parent square root function, defined as follows:

y = sqrt(x).

It's domain, that is, the possible values that the input x can assume, is given as follows:

x ≥ 0.

The graphed function has the same domain as the parent square root function, just with a different range, as it was shifted down, assuming the following format:

y = sqrt(x) - a.

In a translation, the function keeps the same format and orientation, just changing it's position, which is what happened in this case.

The possible translations are:

The other functions would change the domain of the function, hence they are not possible.

The information to solve this problem is given by the image at the end of the answer.

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Answer:

8.97 x 10^6

Step-by-step explanation:

To add in scientific notation, the exponents need to be the same for the bases of 10.  You can either make them both to the exponent of 6 or 5.  It does not matter.  I made them both to the power of 5

(8.6 x10^6 ) + (3.7 x10^5 )

(86 x 10^5) + (3.7x10^5)  

(86 x 3.7) x 10^5

89.7 x 10^5 Rewrite in scientific notation

8.97 x 10^6

The proper subset relationship of the given set A, B and C are in option;

For the given question;

The set are given as;

A = { x | x = 2n + 1, n ∈ R }

This, can be written as for the real number say, n = 0, 1, 2, 3....

A = {1, 5, 7, 9 , 11 ..}

B = { x | x = 4n + 1, n ∈ R }

This, can be written as for the real number say, n = 0, 1, 2, 3....

B = {1, 5, 9, ....}

C = { x | x = 8n + 1, n ∈ R }

This, can be written as for the real number say, n = 0, 1, 2, 3....

C = {1, 9, 17...}

Thus, it is clear from the above relation that,

C contain B and A while B contain some element of A.

Thus, C ⊂ A,  C ⊂ B and B ⊂ A are the correct relationship of the proper subset.

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Answer:

56

Step-by-step explanation:

[tex]\binom{8}{3}=\frac{8!}{5!3!}=56[/tex]

Answer:

12 and 8; 16 and 6; 24 and 4

Step-by-step explanation:

The equation that shows a relationship between t (time in minutes) and d (distance in miles) is d = 0.075t

An equation is an expression that shows the relationship between two or more numbers and variables.

Equations are classified based on degree (value of highest exponents) as linear, quadratic, cubic and so on. Variables can be dependent or independent. Dependent variables depend on other variable while an independent variable do not depend.

The standard form for linear equation is:

y = mx + b

Where m is the slope and b is the y intercept

Let d represent the runners distance after t minutes.

From the table, using the points (12, 0.9) and (16, 1.20):

d - 0.9 = [(1.2 - 0.9)/(16 - 12)](t - 12)

d - 0.9 = 0.075(t - 12)

d = 0.075t

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If guests staying at Marada Inn were asked to rate the quality of their accommodations as being excellent (E), above average (AA), average (A), below average (BA), or poor (P). The frequency is : 1, 8, 6, 3, 2.

Analysis

E = E =1

AA= AA +AA + AA + AA + AA+ AA + AA +AA = 8

A= A + A+ A+ A+ A+ A = 6

BA = BA + BA + BA =3

P = P + P =2

Hence,

Class     frequency

E                     1

AA                  8

A                     6

BA                   3

P                      2

Total               20

Therefore  1, 8, 6, 3, 2 is the frequency .

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Answer:

The triangle shown is an isosceles triangle. It has two equal sides and the base angels are equal.

5x - 7 = 8x - 55

3x = 48

x = 16

Answer: B

Step-by-step explanation: i got it because i

1.

[tex] = (65 \times 12) - 92 + 13 \\ =780 - 92 + 13 \\ = (780 - 92) + 13 \\ = 688 + 13 \\ = 701 [/tex]

2.

[tex] = \frac{4}{2} + 19 -(5 \times 11) \\ = (2 + 19 )- 55 \\ = 21 - 55 \\ = - 34[/tex]

ATTACHED ARE THE SOLUTIONS

Answer:

x = 37

Step-by-step explanation:

∡CAE = 180°

Then:

∡ABC = 180° - (10+8X)

∡ABC = 180 - 10 -8X

∡ABC = 170 -8X

The sum of the internal angles of a triangle results im 180°:

70 + (6x+14) + (170-8x) = 180

70 + 14 + 170 + 6x - 8x = 180

254 - 2x = 180

254 - 180 = 2x

74 = 2x

x = 74/2

x = 37

 ( -4)¹⁹  =  -  274877906944 is standard form of 4 to power -19.

What are power and exponent?

a)  ( -8 )⁰   =   1               ( positive )

b)  ( -4 )⁶    = - 4× -4 × -4 × -4 × -4 × -4 = 4096  ( Positive )

c)  ( -5 )¹²  =  244,1406,25           ( positive )

d)  ( -4)¹⁹  =  -  274877906944   ( Negative)

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Volume of the aquarium is 27 [tex]s^{3}[/tex]

The total number of cubic units that a cube totally fills up is referred to as its volume. An object with six square faces that is three dimensional is called a cube. The amount of space an object takes up in total is known as its volume. Space would be taken up by an object having a larger volume.

Depending on the inputted inputs, many formulas can be used to determine the volume of any cube. You may figure it out by measuring the diagonal of the cube or its sides.

Given that : Carlos bought a cube shaped aquarium with side 3s

As the aquarium have cube shape so the volume will be

Volume of cube is V =[tex]a^{3}[/tex]

a = 3

V = 27 [tex]s^{3}[/tex]

Hence ,Volume of the aquarium is 27 [tex]s^{3}[/tex]

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Explanation:

y% of some value, x:

Convert y to a decimal by dividing y by 100.

Then multiply that decimal value to x.

The original function and its expression derived from rigid transformations are defined below:

In this problem we must use rigid transformations to change the definition of a series of functions. Rigid transformations are transformations applied on functions such that its form is not altered. There are the following transformations, which are used in the five cases:

Horizontal translation

g(x) = f(x - k), translation is to the right for k > 0.

Vertical translation

g(x) = f(x) + k, translation is upward for k > 0

Reflection over the x-axis

g(x) = - f(x)

Reflection over the y-axis

g(x) = f(- x)

Now we proceed to present the image of each function after using a sequence of rigid transformations:

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The warmest recorded temperature than the coldest recorded temperature is 54oF

The given parameters are:

Coldest recorded temperature = 80°F, in Alaska

Warmest recorded temperature = 134°F, in California

Next, we calculate the difference between these temperatures

Difference = Warmest recorded temperature - Coldest recorded temperature

Substitute the known values in the above equation

Difference = 134 - 80

Evaluate

Difference = 54

Hence, the amount of temperature difference is 54oF

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The equation that involves division and addition and has a solution of x=-25 will be 3x - 40 = x + 90

It should be noted that an equation is used to show the relationship between the variables that are given.

The equation that involves division and addition and has a solution of x = -25 will be:

3x - 40 = x - 90

Collect like terms

3x - x = -90 + 40

2x = -50

Divide

x = -50 / 2

x = -25

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Note that:

[tex]\int f(x) \text{ } dx=-e^{-x/5}+C[/tex]

Part (b)

[tex]\int^{1}_{0} f(x) \text{ } dx =-e^{-1/5}+e^{0} \approx 0.1813[/tex]

Part (c)

[tex]\int^{\infty}_{0} f(x) \text{ } dx =1.0000[/tex]

Part (e)

[tex]\int^{6}_{1} f(x) \text{ } dx \approx 0.5175[/tex]

In still air, the rate of the jet is Distance traveled/ time taken

With the jetstream, the rate of jetstream is

A.

The length of the diagonal is given by the Pythagorean theorem therefore

The length of the diagonal is ) units

B.

The area of the square is given by the next formula

where s is the side

s=4

The area of the square is 16 units^2

C.

For the area of the triangle we will use the next formula

where b is the base and h is the height

b=4 units

h=4units

The area of the triangle formed by a diagonal and two of the sides is 8 units^2

D.

For the area of this triangle, we will use the same formula that we use in C. but in this case

b=sqrt(32)/2

h=sqrt(32)/2

We substitute the values

The area of one of these triangles is 4 units^2

ANSWER

A. sqrt(32) units

B.16 units^2

C. 8 units^2

D. 4 units^2

The function, [tex]\displaystyle { p(x) = \frac{ {cos}^{2}x }{sin(2 \cdot x)} }[/tex] has a limit that is undefined (does not exist) as x approaches 0. The correct option therefore option b;

b. As x approaches 0, the limit of p(x) does not exist

The given function is presented as follows;

[tex]\displaystyle { p(x) = \frac{ {cos}^{2}x }{sin(2 \cdot x)} }[/tex]

Required;

The limit of the function as x approaches (zero) 0

Solution;

The limit of a function at a given point within the domain of the function is the value of the function as the function's argument approaches a.

Therefore, the limit of the given function at the point x = 0 is given by the function's value as the argument of the function, x approaches 0.

cos²(0) = 1

sin(2×0) = sin(0) = 0

Therefore;

[tex]\displaystyle { p(x) = \frac{ {cos}^{2}0 }{sin(2 \times 0)} = \frac{ 1 }{0} = \infty}[/tex]

Therefore, the limit of the function does not exist as x approaches 0

The correct option is therefore, option b

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[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{9})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{9}=\stackrel{m}{ \cfrac{1}{5}}(x-\stackrel{x_1}{0})\implies {\Large \begin{array}{llll} y=\cfrac{1}{5}x+9 \end{array}}[/tex]