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3 years ago

10 POINTS WILL MARK BRAINLIEST Find the restriction on y=2x^2-8

1 answer:

7 0

X cannot equal +2, and can't be -1. If you factor it, then you notice you get 2(x+2)(x-2). If you use either one for the x, one will be zero and the won't, and vise-versa.

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NEED RN, WORTH 30! ~ please and thank you! <3

rosijanka [135]

1/4x-4 so the second one

8 0

3 years ago

Jeff purchased 5 items at the same price and 1 item for $2.95. What was the individual price of the 5 items if Jeff's total was

Soloha48 [4]

The answer is A. 3.19, all you have to do it multiply 3.19x5 then add the 2.95 hope this helps

7 0

4 years ago

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

7 0

3 years ago

Ms. Sanches and Mr. Brown went to Chuckie Cheese’s. Ms. Sanches had 567 tokens, and Mr. Brown had 432 tokens. Rounded to the nea

ryzh [129]

Answer:

Rounded to the nearest hundred, Ms. Sanshes had about 200 more tokens than Mr. Brown

Step-by-step explanation:

567=600 (rounded)

432=400 (rounded)

600-400=200 tokens

7 0

3 years ago

Write a quadratic function to model the vertical motion for each situation, given h(t) equals negative 16t squared plus v0t+h0

Mashcka [7]

Answer:

hmax = 194 ft

The maximum height is 194 ft

Step-by-step explanation:

According to the given equation for the model of the vertical motion. The height at any point in time can be written as;

h(t) = -16t^2 + v0t + h0 .......1

Where;

h(t) = height at time t

t = time

v0 = initial velocity = 96 ft/s

h0 = initial height = 50 ft

To determine the maximum height we need to differentiate the equation 1 to find the time at which it reaches maximum height;

At the highest point/height h' = dh/dt = 0

h'(t) = -32t +v0 = 0

-32t + v0 = 0

t = v0/32

t = 96/32

t = 3 s

At t=3 it is at maximum height.

The maximum height can be derived from equation 1;

Substituting the values of t,v0,h0 into equation 1;

h(t) = -16t^2 + v0t + h0 .......1

hmax = -16(3)^2 + 96(3) + 50 = 194 ft

hmax = 194 ft

The maximum height is 194 ft

5 0

4 years ago