Are the ratios 16:8 and 4:2 equivalent?

While passing a slower car on the highway, i accelerate uniformly from 12 m/s to 24 m/s in a time of 10.0s. How far do you trave

Taya2010 [7]

You say that there is uniform acceleration so:

vf-vi=at (final velocity minus initial velocity is equal to acceleration times time)

We know vf, vi, and t so we can solve for acceleration:

24-12=a10

12=10a

a=1.2

That is the acceleration, we will need to integrate with respect to time twice...

v=⌠a dt

v=at+vi , we know a=1.2m/s^2 and vi=12m/s

v=1.2t+12,

x=⌠1.2t+12 dt

x=1.2t^2/2+12t+xo, we can just let xo=0 for this problem...

x(t)=0.6t^2+12t

Now we know that this acceleration lasts for 10 seconds so the distance traveled in that time is:

x(10)=0.6(10^2)+12(10)

x(10)=60+120

x(10)=180 meters

6 0

3 years ago

At the city museum, child admission is

Irina18 [472]

You have to multiply ......... and use 186 and divide. i think.

5 0

3 years ago

What is the equation of the line parallel to the given line with an X intercept of 4

mixas84 [53]

You haven't shared the given line, so all I can do here is to invent a line and then show you how to write the equation of a new line which is parallel to mine and which has an x-intercept of 4.

My invented line: y = (2/3)x + 3

The new line MUST have the same slope: m = 2/3.

Then y = mx + b becomes y = (2/3)x + b. Find the x-intercept by setting y = 0 and solving for x: (2/3)x = 0 - b. Now replace x with 4 and find b:

-b = (2/3)(4) = 8/3. Then b = -8/3, and the new line is

y = (2/3)x - 8/3.

3 0

4 years ago

Read 2 more answers

Find the area.<br> 11 cm<br> 8 cm<br> 7 cm

HACTEHA [7]

66.5cm

Split the shape into a square and triangle and find the area separately

Square: 8*7=57
Triangle: 7*3/2 = 10.5
=66.5

7 0

3 years ago

Read 2 more answers

Given f(x) = 4x^4 find f^-1(x) Then state whether f^-1(x) is a function.

sdas [7]

Answer:

$f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$
$f^{-1}(x) \quad\text{is not a function}$

Step-by-step explanation:

To find the inverse function, solve for y:

$x=f(y)\\\\x=4y^4\\\\\dfrac{x}{4}=y^4\\\\\pm\sqrt[4]{\dfrac{x}{4}}=y\\\\f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$

f(x) is an even function, so f(-x) = f(x). Then the inverse relation is double-valued: for any given y, there can be either of two x-values that will give that result.

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A function is single-valued. That means any given domain value maps to exactly one range value. The test of this is the "vertical line test." If a vertical line intersects the graph in more than one point, then that x-value maps to more than one y-value.

The horizontal line test is similar. It is used to determine whether a function has an inverse function. If a horizontal line intersects the graph in more than one place, the inverse relation is not a function.

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Since the inverse relation for the given f(x) maps every x to two y-values, it is not a function. You can also tell this by the fact that f(x) is an even function, so does not pass the horizontal line test. When f(x) doesn't pass the horizontal line test, f^-1(x) cannot pass the vertical line test.

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The attached graph shows the inverse relation (called f₁(x)). It also shows a vertical line intersecting that graph in more than one place.

5 0

3 years ago