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3 years ago

Order the numbers from least to greatest. 8, 0, -8, -3, 5

Mathematics

1 answer:

Kitty [74]3 years ago

7 0

Answer:

-8,-3,0,5,8

Step-by-step explanation:

Number the the negatives from greatest to least first and continue to number the positives from least to greatest.

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Anyone Wanna help thanks

Oksi-84 [34.3K]

Answer:

1. 8

2. 4

3a. -6 and -4

3b. -1/2 and -5

Step-by-step explanation:

3a explanation:

Factor x^2+10x+24

(x+4)(x+6)

Set the factors equal to 0

x + 4 = 0

x + 6 = 0

-4 and -6

3b.

Factor left side of equation.

(2x+1)(x+5)=0

Set factors equal to 0.

2x + 1 = 0 or

x + 5 = 0

x = −1/2

x = -5

-5 and -1/2

6 0

3 years ago

Write an explicit formula for the sequence -13,-6,1,8, ...

dimulka [17.4K]

Answer:

7n-20 I believe

Step-by-step explanation:

hope this helps

4 0

2 years ago

Read 2 more answers

Which of the following is the solution to |x-13|<18

serious [3.7K]

Rewrite the inequality without the absolute value

-18 < x - 13 < 18

Add 13 to the whole equation

-18 + 13 < x < 18 + 13

Simplify

3 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.