Somebody help me please

Answer:

The angle that will be congruent to angle 4 is :

                             Angle 1

Step-by-step explanation:

It is given that:

angle 4 and angle 5 are complements.

Also, angle 1 and angle 5 are complements.

Congruent complementary Theorem--

It states that if two angles are complementary to the same angle, then the two angles are congruent to each other.

Here both angle 1 and angle 4 are complementary to the same angle i.e. angle 5.

Answer:

40.2 in²

Step-by-step explanation:

h*b/2

h= 6.7

b= 12

a. The third-order Taylor polynomial,  P3(x) = f(xo) + f'(xo)(x - xo) + (f''(xo)(x - xo)^2)/2 + (f'''(xo)(x - xo)^3)/6.

b. The polynomial P3(x) obtained in part a can be used to approximate f(0.1).

c. The polynomial P3(x) obtained in part a can be used to approximate the integral of (1+x)/(1+x^2) from 0 to 0.1.

a. To calculate the third-order Taylor polynomial P3(x) about xo for f(x), we need to find the values of f(x), f'(x), f''(x), and f'''(x) at x = xo. Once we have these values, we can use the formula: P3(x) = f(xo) + f'(xo)(x - xo) + (f''(xo)(x - xo)^2)/2 + (f'''(xo)(x - xo)^3)/6. Plugging in the values of f(xo), f'(xo), f''(xo), and f'''(xo) will give us the third-order Taylor polynomial.

b. The polynomial P3(x) obtained in part a can be used to approximate the value of f(0.1). We can substitute x = 0.1 into P3(x) to obtain the approximation.

c. Similarly, the polynomial P3(x) obtained in part a can be used to approximate the integral of (1+x)/(1+x^2) from 0 to 0.1. We can evaluate the polynomial P3(x) at x = 0.1 and substitute the result into the integral expression.

In summary, the third-order Taylor polynomial P3(x), about xo, for f(x) is calculated using the formula involving the values of f(xo), f'(xo), f''(xo), and f'''(xo). This polynomial can then be used to approximate the value of f(0.1) and the integral of a given function.

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Answer:

C NO TOO HOT OR TOO COLD

Step-by-step explanation:

Answer:

the answer is c

Step-by-step explanation:

Brainliest plz

option D: Horizontal scale: 1 unit = 10, Vertical scale: 1 unit = 10 is the correct answer.

To determine an appropriate horizontal scale and vertical scale for the viewing window, one must analyze the problem, the graph, and the data presented in the problem. The appropriate horizontal scale and vertical scale for the viewing window is given by option D, that is Horizontal scale: 1 unit = 10, Vertical scale: 1 unit = 10. The horizontal scale and vertical scale refer to the scale of the x-axis and y-axis of the graph. These scales are used to represent data in a graphic format that can be easily understood by readers. The horizontal scale is used to determine the value of the x-axis in the graph. The vertical scale is used to determine the value of the y-axis in the graph. The scale should be chosen based on the range of values in the graph. When the values in the graph are too small or too large, the scale must be adjusted accordingly. In this case, since the values of horizontal and vertical scale are large, the scale must be adjusted accordingly. Therefore, option D is the correct answer.

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The size of the horizontal and vertical scales should be chosen so that the graph fits in the viewing window. So, that is why the answer is "b. Horizontal scale: 1 unit = 5, Vertical scale: 1 unit = 10."

When graphing a function, it is important to choose an appropriate horizontal scale and vertical scale for the viewing window. The answer is "b. Horizontal scale: 1 unit = 5, Vertical scale: 1 unit = 10."Why is the answer (b) Horizontal scale: 1 unit = 5, Vertical scale: 1 unit = 10?The answer to this question is based on the graph. It is more appropriate to use (b) horizontal scale: 1 unit = 5, vertical scale: 1 unit = 10. In this situation, the x-axis should have a horizontal scale of 1 unit = 5 so that the x-axis can fit within the viewing window. Meanwhile, the y-axis should have a vertical scale of 1 unit = 10 so that the entire graph can fit in the viewing window.

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The exact value of A in the general solution y = A + cx is 75

From the question, we have the following parameters that can be used in our computation:

y = A + cx

The differential equation is given as

y - xy' = 75

When y = A + cx is differentiated, we have

y' = c

So, we have

y - xc = 75

Recall that

y = A + cx

So, we have

A + cx - xc = 75

Evaluate the like terms

A = 75

Hence, the value of A in the general solution is 75

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Answer:

Rewrite the question correctly! it's totally absurd

Answer:

y

Step-by-step explanation:

y needs to provide a number

Answer:

-4, -7, -2, -5, 0, -3, 2, -1

Step-by-step explanation:

the pattern is -3, +5.

i hope this helps :)

Answer:

4(x+5)

Step-by-step explanation:

according to your question

Answer:

(5,-1)

Step-by-step explanation:

Answer:

$400

Step-by-step explanation:

Let the initial total cost be represented = $X.

Hence, for 20 family members

= $ x/20 each .......1

Now that they are 25 members

= $x/25 each.......2

Each in the original group which is equation 1 owed $4 less.

That is,

x/20 - 4 ........ 3

Now equate 2 and 3

X/20 - 4/1 = X/25

Taking the LCM of both sides

X-80/20 = X/25

Cross multiply both sides

25( x-80 ) = 20x

25x-2000 = 20x

Collecting like terms

25x-20x = 2000

5x = 2000

Divide through by 5

X = 2000/5

X = 400

The total cost for the catered meal is $400.

To check if it was correct

Put 400 into equation 1 and 2 and then minus whatever you get from each other, you will get $4

From equation 1

400/20 = $20

From equation 2

400/25 =$16

$20-$16 = $4

Thanks

Answer:

-5.75736     or    [tex]\frac{1}{-10+3√2}[/tex]

Step-by-step explanation

Answer:

Multiply the volume of the cone by 3

Step-by-step explanation:

Required

Volume of a cylinder from a cone of the same dimension

The volume of a cone is:

[tex]V_1 = \frac{1}{3}\pi r^2h[/tex]

The volume of a cylinder is:

[tex]V_2 = \pi r^2h[/tex]

Recall that:

[tex]V_1 = \frac{1}{3}\pi r^2h[/tex]

Split:

[tex]V_1 = \frac{1}{3}*[\pi r^2h][/tex]

Substitute: [tex]V_2 = \pi r^2h[/tex]

[tex]V_1 = \frac{1}{3}*V_2[/tex]

Make V2 the subject

[tex]V_2 =3V_1[/tex]

This implies that, we simply multiply the volume of the cone by 3

Answer:

3,454 cm³

Step-by-step explanation:

This is a cylinder so the volume formula is [tex]V=\pi r^{2} h[/tex]

The problem gives us the diameter (20 cm) and the height (11 cm). Since the formula uses the radius instead of the diameter, we have to divide the diameter by 2.

20 ÷ 2 = 10

The radius is 10 cm

Now, we plug our height and radius into the formula to find the volume.

[tex]V=\pi r^{2} h\\V=(\frac{22}{7})(10)^{2}(11)\\V=(\frac{22}{7})(100)(11)\\[/tex]

V≈3,454

Answer:

121 centimeters ²

Step-by-step explanation:

Area of a square = length ²

Each side of a square = 5x

find the area of the square when x = 2.2 centimeters

Area of a square = length ²

= (5x)²

= 5x * 5x

= 5(2.2) * 5(2.2)

= 11 * 11

= 121 centimeters ²

Area of a square = 121 centimeters ²

The value of a in the given function is 4

To find the value of a in the equation f(g(x)) = (1 - x²)³, we need to substitute the function g(x) = a + x² into the function f(x) = (5 - x)³.

Replacing x in f(x) with g(x), we have:

f(g(x)) = (5 - g(x))³.

Substituting g(x) = a + x², we get:

f(g(x)) = (5 - (a + x²))³.

f(g(x)) = (5 - a - x²)³.

Comparing this expression with (1 - x²)³, we can equate the corresponding terms:

5 - a - x² = 1 - x².

5 - a = 1.

a = 5 - 1

a = 4

Therefore, the value of a is 4.

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X denote the amount of time a book on two-hour reserve is actually checked out, and for the given cdf the following are the answers for the questions asked

(a) P(X ≤ 1) ≈ 0.0625

(b) P(0.5 ≤ X ≤ 1) ≈ 0.0469

(c) P(X > 1.5) ≈ 0.8594

(d) Median checkout duration ≈ 2.828

(e) Density function f(x) is defined as:

f(x) = 0 for x < 0

f(x) = x/8 for 0 ≤ x ≤ 4

f(x) = 0 for x > 4

(f) Expected value E(X) ≈ 2.667

(g) Variance V(X) ≈ 0.889, Standard Deviation σ(X) ≈ 0.943

(h) Expected charge E[h(X)] = 8

In probability theory, a cumulative distribution function (CDF) provides information about the probabilities of certain events occurring in a random variable. In this scenario, let's consider a book that is on a two-hour reserve and denote the amount of time it is checked out as X. We are given the CDF of X and we will use it to calculate various probabilities and statistics related to the checkout duration of the book.

The given CDF is as follows:

F(x) = 0 for x < 0

F(x) = x²/16 for 0 ≤ x ≤ 4

F(x) = 1 for x > 4

(a) P(X ≤ 1):

To calculate this probability, we need to find F(1) since F(x) represents the cumulative probability up to x. From the given CDF, we see that F(x) = x²/16 for 0 ≤ x ≤ 4. Substituting x = 1 into the equation, we get:

F(1) = (1²)/16 = 1/16.

(b) P(0.5 ≤ X ≤ 1):

To calculate this probability, we need to find F(1) - F(0.5) since F(x) represents the cumulative probability up to x. From the given CDF, we have F(0.5) = (0.5²)/16 = 1/64 and F(1) = (1²)/16 = 1/16. Therefore,

P(0.5 ≤ X ≤ 1) = F(1) - F(0.5) = (1/16) - (1/64) = 3/64.

(c) P(X > 1.5):

To calculate this probability, we need to find 1 - F(1.5) since F(x) represents the cumulative probability up to x. From the given CDF, we have F(1.5) = (1.5²)/16 = 9/64. Therefore,

P(X > 1.5) = 1 - F(1.5) = 1 - (9/64) = 55/64.

(d) Median checkout duration:

The median is the value that divides the distribution into two equal parts, meaning that half of the checkouts are below this value and half are above it. We need to solve the equation F(median) = 0.5. From the given CDF, we have:

F(median) = 0.5

0.5 = (median²)/16

Solving for the median, we get median = √(8) ≈ 2.828.

(e) Density function f(x):

The density function f(x) represents the derivative of the cumulative distribution function F(x). To obtain f(x), we differentiate the given CDF:

f(x) = F'(x)

For x < 0, f(x) = 0 since F(x) is constant in that range.

For 0 ≤ x ≤ 4, we have F(x) = x²/16.

Differentiating with respect to x, we get:

f(x) = d/dx (x²/16) = (2x)/16 = x/8.

For x > 4, f(x) = 0 since F(x) is constant in that range.

Therefore, the density function f(x) is:

f(x) = 0 for x < 0

f(x) = x/8 for 0 ≤ x ≤ 4

f(x) = 0 for x > 4

(f) Expected value E(X):

The expected value of a random variable X is a measure of its average value. To calculate E(X), we integrate the product of x and the density function f(x) over the entire range of X:

E(X) = ∫[x * f(x)] dx

For x < 0 and x > 4, f(x) = 0, so we only need to consider the interval 0 ≤ x ≤ 4:

E(X) = ∫[x * (x/8)] dx

= (1/8) ∫[x²] dx (integrating x²)

= (1/8) * (x³/3) + C (integrating x²)

= (1/24) * (x³) + C

Evaluating this expression from x = 0 to x = 4, we get:

E(X) = (1/24) * (4³) - (1/24) * (0³)

= 64/24

= 8/3

≈ 2.667

(g) Variance V(X) and Standard Deviation σ(X):

Variance is a measure of the spread or dispersion of a random variable. To calculate V(X), we need to calculate the second moment E(X²) and subtract the square of the expected value [E(X)]². The standard deviation σ(X) is the square root of the variance.

E(X²):

E(X²) = ∫[x² * f(x)] dx

For x < 0 and x > 4, f(x) = 0, so we only need to consider the interval 0 ≤ x ≤ 4:

E(X²) = ∫[x² * (x/8)] dx

= (1/8) ∫[x³] dx (integrating x³)

= (1/8) * (x⁴/4) + C (integrating x³)

= (1/32) * (x^4) + C

Evaluating this expression from x = 0 to x = 4, we get:

E(X²) = (1/32) * (4⁴) - (1/32) * (0⁴)

= 256/32

= 8

V(X):

V(X) = E(X²) - [E(X)]²

= 8 - (8/3)²

= 8 - 64/9

= 8 - 7.111

≈ 0.889

Standard deviation:

σ(X) = √(V(X))

= √(0.889)

≈ 0.943

(h) Expected charge E[h(X)]:

Given the function h(X) = X², we want to calculate the expected value of h(X). This can be done by finding E[h(X)] = E(X²).

From the previous calculations, we know that E(X²) = 8. Therefore, the expected charge is E[h(X)] = 8.

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Area= 1/2 (a+b)h

1/2(1in. +3in.)1in.

1/2(4in.)1in.

2in.× 1in.

2in. Therefore the area is 2in.

Trick question! All are rational numbers.

..................................................................................

Answer:

I got about 28.01 ft you might want to round or something

hope this helped

The required stopping distance of the car is 17.24 metes.

A child appears to be running into the street ahead. It takes 2.3 seconds for the driver to react and begin to brake, but this time at a rate of -7.5 m/s2. What is the stopping distance for the car in this situation is to be determined.

Speed is ratio of distance to the time. speed = distance / time.

The intial speed of the driver is 11m/s

Now total stopping distance = thinking distance + break applying distance
                                           =  v*t  + u²/2a
                                           =  11 * 2.3 - 11²/2*7.5
                                           = 17.24 meters,

Thus, the required stopping distance of the car is 17.24 metes.

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Answer:

it's the third option

Step-by-step explanation:

I think I'm pretty sure

hope this helped ;)

Answer:

0

Step-by-step explanation:

HOPE I HELPED!

Answer:

Yes lol

Step-by-step explanation:

Well the app says it helps people learn the answers but when I'm in a test i just look at letters:)

Answer:

I mean i'm the type to give a long winded answer. But i guess it depends on the context. Sometimes people ask for an explanation yk? Like most of those ppl who ask questions for History and English need a written response.

Answer:16

Step-by-step explanation:

33-9=24

24/1.50=16

(a) The Z-Score is 1.78,

(b) The two sample-sizes are 5070 and 1268,

(c) The researchers should choose the 2% margin-of-error.

Part (a) : To find Z, we use the confidence-level (CL) to find the corresponding Z-score.

The confidence-level (CL) is given as 92.5%, which corresponds to an area of 0.925 under the standard normal-distribution curve. Since the remaining area on both tails is (1 - 0.925) = 0.075, we divide this value by 2 to get the area for one tail: 0.075/2 = 0.0375,

The "Z-score" that corresponds to area of 0.0375 in one tail. The Z-score is approximately 1.78,

Part (b) : To determine the two sample sizes for the 1% and 2% margin of error, we use the formula,

Sample size (n) = (Z² × σ²)/(E²),

For a 1% margin-of-error (E = 0.01),

We have,

n₁ = (1.78² × 0.4²)/(0.01²),

n₁ ≈ 5070

For a 2% margin of error (E = 0.02),

We have,

n₂ = (1.78² × 0.4²)/(0.02²),

n₂ ≈ 1268

Part (c) : The researchers should choose the margin-of-error that results in the least possible number of respondents since they have a limited budget.

Comparing the two sample-sizes, we find that sample-size for 2% margin of error (n₂ ≈ 1268) is smaller than the sample-size for the 1% margin of error (n₁ ≈ 5070).

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Answer:

v = -3/2

Step-by-step explanation:

6v = -9

v = -9/6

v = -3/2