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3 years ago

What is the function rule represented by the following table?

Mathematics

2 answers:

Hunter-Best [27]3 years ago

3 0

Answer:

D. y = -4x + 3

Step-by-step explanation:

x y

0 3 -4(0) + 3 = 3

1 -1 -4(1) + 3 = -1

2 -5 -4(2) + 3 = -5

eduard3 years ago

3 0

D. y= -3x Hope this helps you on your test

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All 231 students in the Math Club went on a field trip. Some students rode in vans which hold 7 students and some students rode

riadik2000 [5.3K]

answer : 8 vans and 7 buses

7 0

4 years ago

Read 2 more answers

Solve 6x-54=6(x-9)<br><br> show work please

Stolb23 [73]

Answer:infinite

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

6x−54=6(x−9)

6x+−54=(6)(x)+(6)(−9)(Distribute)

6x+−54=6x+−54

6x−54=6x−54

Step 2: Subtract 6x from both sides.

6x−54−6x=6x−54−6x

−54=−54

Step 3: Add 54 to both sides.

−54+54=−54+54

0=0

5 0

3 years ago

A rock thrown vertically upward from the surface of the moon at a velocity of 36m/sec reaches a height of s = 36t - 0.8 t^2 met

Verdich [7]

Answer:

a. The rock's velocity is $v(t)=36-1.6t \:{(m/s)}$ and the acceleration is $a(t)=-1.6 \:{(m/s^2)}$

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

Velocity is defined as the rate of change of position or the rate of displacement. $v(t)=\frac{ds}{dt}$
Acceleration is defined as the rate of change of velocity. $a(t)=\frac{dv}{dt}$

The rock's velocity is the derivative of the height function $s(t) = 36t - 0.8 t^2$

$v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t$

The rock's acceleration is the derivative of the velocity function $v(t)=36-1.6t$

$a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6$

b. The rock will reach its highest point when the velocity becomes zero.

$v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5$

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

$s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405$

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = $\frac{405}{2} =202.5 \:m$

To find the time it reach half its maximum height, we need to solve

$36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41$

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

$36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45$

The rock is aloft for 45 seconds.

5 0

4 years ago

Please help! It’s not too hard! Will mark brainliest!

Yuki888 [10]

Step-by-step explanation:

In the first part, he was travelling with twice the speed, or Vs, for distance

48 - d, for 2 hours.

speed = distance/time

distance = speed * time

First part :

48 - d = 2V * 2

⇒d + 4V = 48

Second part:

d = V * 2

⇒d = 2V

Substitute 2s for d in equation 1.

2s + 4s = 48

6s = 48

s = 8

The speed in the second part is 8 mph.

<u>The speed in the first part is 2s = 2(8) = 16</u>

I hope this helps

7 0

3 years ago

Which expression is equivalent to \root(3)(x^(5)y)

san4es73 [151]

Answer:

$\sqrt[3]{x^{5}y} = x^{\frac{5}{3}}y^\frac{1}{3}$

Step-by-step explanation:

Given

$\sqrt[3]{x^{5}y}$

Required

The equivalent expression

We have:

$\sqrt[3]{x^{5}y}$

Rewrite as an exponent

$\sqrt[3]{x^{5}y} = (x^{5}y)^\frac{1}{3}$

Open bracket

$\sqrt[3]{x^{5}y} = x^{5*\frac{1}{3}}y^\frac{1}{3}$

$\sqrt[3]{x^{5}y} = x^{\frac{5}{3}}y^\frac{1}{3}$

6 0

3 years ago