carla travels on a high speed train for 30 minutes. If she travels 160 km, how many kilometers will she travel in one hour


plz help

Answer:

A. f(x) = 3*(1/2)^x

Step-by-step explanation:

We know that, a function can be stretched or shrinked both horizontally and vertically.

Now, according to our question we are required to look at the vertical stretch of an exponential function.

The general form for a vertical stretch of a function f(x) is k*f(x) where k>1.

So, we compare this form with the options provided.

We see that in option A the exponential function is multiplied by 3 and so the function will be stretched vertically.

Hence, option A is correct.

Answer:

A. the volleyball reached it's maximum height of 3 sec

B. The maximum height of the vollybal was 23 ft

E. The vball was served from a height of 5 ft

Step-by-step explanation:

h(t) = -2 (t-3)^2 +23

Given equation is in the form of [tex]f(x)= a(x-h)^2 + k[/tex]

(h,k) is the vertex

Now we compare f(x) with h(t)

h(t) = -2 (t-3)^2 +23

h = 3  and k = 23

Vertex is (3,23)

h=3 . this means the volleyball reaches its maximum height in 3 seconds

k = 23. this means the volleyball reaches the maximum height of 23 ft

When ball reaches the ground the height becomes 0. so plug in 0 for h(t) and solve for t

0= -2 (t-3)^2 +23

Subtract 23 on both sides

-23 = -2(t-3)^2

Divide both sides by -2

[tex]\frac{-23}{-2} = (t-3)^2[/tex]

Take square root on both sides

[tex]+-\sqrt{\frac{23}{2}}= t-3[/tex]

Add 3 on both sides

[tex]+-\sqrt{\frac{23}{2}}+3= t[/tex]

We will get two value for t

t=-0.39  and t= 6.39

So option C is not correct

Given h(t) is a quadratic function not exponential

To find initial height we plug in 0 for x and find out h(0)

h(0) = -2 (0-3)^2 +23 = -2(-3)^2 + 23= -18+ 23= 5

The volleyball was served from a height of 5 ft

The acceleration of the body at t = 3 seconds is approximately  [tex]\( -58.58785 \, \text{m/s}^2 \).[/tex]

To find the acceleration of the body at ( t = 3 ) seconds, we need to find the second derivative of the displacement function [tex]\( x(t) \)[/tex] with respect to time t.

Given that[tex]\( x(t) = a \sin(\omega t + \phi) \)[/tex] , where [tex]\( a = 5 \) m, \( \omega = 3.444 \) rad/s, and \( \phi = 1.0472 \)[/tex] rad, the acceleration a(t) is the second derivative of of x(t) with respect to t:

First, let's find the first derivative of of x(t) with respect to t:

[tex]\[ v(t) = \frac{dx}{dt} = a \omega \cos(\omega t + \phi) \][/tex]

Now, let's find the second derivative of x(t) with respect to t:

[tex]\[ a(t) = \frac{d^2x}{dt^2} = -a \omega^2 \sin(\omega t + \phi) \][/tex]

Now, substitute the given values:

[tex]\[ a(t) = -5 \times (3.444)^2 \sin(3.444 \times 3 + 1.0472) \][/tex]

Now, calculate:

[tex]\[ a(t) = -5 \times (3.444)^2 \sin(10.332 + 1.0472) \]\[ a(t) = -5 \times (3.444)^2 \sin(11.3792) \][/tex]

Now, compute:

[tex]\[ a(t) = -5 \times (11.858736) \sin(11.3792) \]\[ a(t) \approx -5 \times (11.858736) \times 0.97989 \]\[ a(t) \approx -58.58785 \, \text{m/s}^2 \][/tex]

Therefore, the acceleration of the body at t = 3 seconds is approximately  [tex]\( -58.58785 \, \text{m/s}^2 \).[/tex]

Final answer:

To find the number of members for any year on the video sharing website, use the recursive rule with a base case of 20,000 initial members and apply the recursive step Mₙ = 0.75 × Mₙ₋₁ + 10,000 for each subsequent year.

Explanation:

The situation described can be modeled with a recursive rule where each year's membership is based on the previous year's membership.

To write such a rule, we'll use Mn to denote the number of members in year n, and Mn-1 to denote the number of members in the previous year.

The recursive rule is as follows:

This means that for any year n, the number of members is equal to 75% of the previous year's members plus an additional 10,000 new members.

To determine the probability of finding at least one defective bulb in the lot of 100 bulbs, one must calculate the complementary event, picking no defective bulbs, and subtract its probability from 1.

This problem involves Probability and Combinatorics, fundamental branches in Mathematics to calculate the possible outcomes in an event. Here, to calculate the probability of finding at least one defective bulb, we can begin by calculating the complementary event, that is, finding no defective bulbs in a sample.

The total number of ways to pick 10 bulbs from 100 is the combination C(100, 10). Likewise, the total number of ways to pick 10 non-defective bulbs from the 95 non-defective bulbs in the lot is C(95, 10). The probability of picking 10 non-defective bulbs then is C(95, 10)/C(100, 10).

Since the event of finding 'at least one defective bulb' is the complementary of 'finding no defective bulbs', we can subtract this probability from 1. So, the probability of finding at least one defective bulb is 1 - (C(95, 10)/C(100, 10)).

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The volume of the cylinder is; 128π in3

Base area(πr^2)*height

Π = 3.14

Therefore; 3.14 * 4^2 * 8

3.14 * 128

= 401.92 in3

The total amount paid = $3840

"A = P(1 + rt)

Where A = Total Amount (principal + interest)

P = Principal Amount

I = Interest Amount

r = Rate of Interest in decimal

R = Rate of Interest as a percent

[tex]r=\frac{R}{100}[/tex]

t = Time Period"

For given question,

P = $3200

t = 4 years

R = 5%

[tex]\Rightarrow r=\frac{5}{100}\\\\ \Rightarrow r=0.05[/tex]

Using the formula of simple interest,

[tex]A=P(1+rt)\\\\A=3200(1+(0.05\times 4))\\\\A=3200(1+0.2)\\\\A=3200\times 1.2\\\\A=\$ 3840[/tex]

Therefore, the total amount paid = $3840

Learn more about simple interest here:

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Angle a and angle b are pairs of angles called the alternate angles. Alternate angles are pair of angles that is formed by two parallel lines. These two angles are not adjacent to each other. They are located on the opposite sides of the line that intersects the parallel lines.

Final answer:

To find the limit of f(x) as x approaches 3, factor the numerator and denominator as differences of squares, cancel common terms, and substitute x = 3 into the simplified fraction to get the limit, which is 4.5.

Explanation:

The student's schoolwork question involves calculating the limit of a function as x approaches a certain value. Specifically, the function in question is f(x) = (x^3 - 27) / (x^2 - 9) and the limit to be computed is limx → 3 f(x). To solve this, first recognize that directly substituting x = 3 would result in a 0/0 indeterminate form. To find the limit, we'll use algebraic manipulation.

Since both the numerator and the denominator are differences of squares, factoring them would give (x-3)(x^2+3x+9) for the numerator and (x-3)(x+3) for the denominator. The x-3 terms cancel out, leaving us with x^2+3x+9 over x+3. Now, substituting x = 3 into the simplified fraction will provide the value of the limit.

Calculating the limit step by step:

Original function: f(x) = (x^3 - 27) / (x^2 - 9)

Factor both numerator and denominator: f(x) = [(x-3)(x^2+3x+9)] / [(x-3)(x+3)]

Cancel out (x-3) terms: f(x) = (x^2+3x+9) / (x+3)

Substitute x = 3: f(3) = (3^2+3*3+9) / (3+3) = (9+9+9) / 6 = 27 / 6 = 4.5

The differential of the function v = 2y cos(xy) is found using the product rule and the chain rule, yielding dv = 2 cos(xy) dy - 2y sin(xy) (xdy + ydx), which accounts for changes with respect to both x and y.

To find the differential of the function v = 2y cos(xy), we'll need to apply the product rule and the chain rule. The product rule states that the differential of a product of two functions is d(uv) = u dv + v du. The chain rule is used when differentiating composite functions, and it implies that d/dx (f(g(x))) = f'(g(x)) * g'(x).

Applying the product rule, we get:

Combining these, the differential dv becomes:

dv = 2 cos(xy) dy - 2y sin(xy) (xdy + ydx).

This equation represents the total differential of the function v with respect to both x and y.

For implicit differentiation examples like in equation In(xy) = x2y3 or systems of differential equations, understanding and applying the product rule, chain rule, and implicit differentiation are essential to calculate derivatives and solve the equations.

To write the decimal number 0.8(repeating) as a fraction, use the concept of an infinite geometric series and convert it to an equation. Multiply both sides of the equation by 10 to eliminate the decimal point and subtract the original equation to eliminate the repeating part. Finally, divide by 9 to get the equivalent fraction form.

To write the decimal number 0.8(repeating) as a fraction, we can use the concept of an infinite geometric series. Let's assume the repeating decimal as x: x = 0.8(repeating). Now, multiply both sides of the equation by 10 to remove the decimal point: 10x = 8(repeating). Next, subtract the original equation from this new equation to eliminate the repeating part: 10x - x = 8(repeating) - 0.8(repeating). Simplifying these expressions gives us: 9x = 7.2. Finally, divide both sides of the equation by 9: x = 7.2/9. Therefore, the equivalent fraction form of 0.8(repeating) is 7.2/9.

We know that [tex] \boldsymbol{A^2}=\boldsymbol{A\times A} [/tex]

We also know that [tex] \boldsymbol{I}=\begin{bmatrix}
1 &0 \\
0&1
\end{bmatrix} [/tex]

Now, it has been given to us that [tex] \boldsymbol{A^2}=\boldsymbol{I} [/tex]

Therefore, we will have to find the correct [tex] \boldsymbol{A} [/tex] from the given options and we find that when:

[tex] \boldsymbol{A}=\begin{bmatrix}
3 &-2 \\
4&-3
\end{bmatrix} [/tex]

then [tex] \boldsymbol{A^2}=\boldsymbol{A\times A}=\begin{bmatrix}
3 &-2 \\
4&-3
\end{bmatrix}\times \begin{bmatrix}
3 &-2 \\
4&-3
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix} [/tex]

Therefore, from the above given options, Option C is the correct option.

Therefore, [tex] \boldsymbol{A}=\begin{bmatrix}
3 & -2\\
4 & -3
\end{bmatrix} [/tex] is the correct answer.

Answer:

C. [3, -2]

    [4, -3]

Step-by-step explanation:

PLATOOOO

Answer:

36 cm^2

Step-by-step explanation: