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3 years ago

Find the zeros K(x) = -x²+ 2x + 15 EXPLAIN EVERY STEP!

1 answer:

5 0

Take out - 1
k(x) = -1(x^2 - 2x - 15) Now factor what's inside the brackets. The numbers you pick must add to - 2 and multiply to -15.
When you multiply to -15 you should note that one of the numbers is plus and the other is minus.
(x - )(x + ) is the way it looks.
They differ by 2 and the "largest" one is minus. That means your answer is -5 and +3
k(x) = - (x - 5)(x + 3)
Now for the zeros.
x - 5 can be a zero
x - 5 = 0
x = 5

x + 3 = 0
x = - 3

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On a trip to Griffith Observatory, Dave rode his bicycle six more than twice as many miles in the afternoon as in the morning. I

podryga [215]

Answer:

Dave rode his bicycle 17 miles in the morning

Dave rode his bicycle 40 miles in the afternoon

Step-by-step explanation:

* Lets study the information to solve the question

- Dave rod his bicycle in the morning and again in the afternoon

- Six more than twice as many miles in the afternoon as in the morning

means the distance in the afternoon is 6 more than twice the distance

in the morning

- The entire trip was 57 miles means the total distance in the morning

and in the afternoon was 75

* To solve the question let the distance in the morning is x miles

∵ The distance in the morning = x miles

∵ The distance in the afternoon is 6 more than twice the distance

in the morning

∴ The distance in the afternoon = 2x + 6 miles

∵ The distance in entire trip = 57 miles

- The distance in the morning and the distance in the afternoon = 57 miles

∴ x + (2x + 6) = 57 ⇒ simplify

∴ 3x + 6 = 57 ⇒ subtract 6 from both sides

∴ 3x = 51 ⇒ divide both sides by 3

∴ x = 17 miles

∵ The distance in the morning is x miles

∴ Dave rode his bicycle 17 miles in the morning

∵ The distance in the afternoon = 2x + 6 miles

- Substitute the value of x

∴ The distance in the afternoon = 2(17) + 6 = 34 + 6 = 40 miles

∴ Dave rode his bicycle 40 miles in the afternoon

4 0

4 years ago

Amy read 8 more books than ben read. how many books did ben read

zaharov [31]

Answer:

(x+8)-x

Step-by-step explanation:

if amy is 8 more than ben

we dont know ben, so he is x

you have to subtract ben from amy

so amy is 8 more than ben, you have to add 8 to ben, who is x

6 0

3 years ago

Read 2 more answers

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.