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2 years ago

6.4 x 10 to the power of 17

Mathematics

2 answers:

lina2011 [118]2 years ago

8 0

6.4 x 10 = 64, however, that is not the answer.

64^17 would be 64 multiplied upon itself 17 times.

The answer you should submit should be 64^17 since the number it creates is too big to calculate

Darya [45]2 years ago

4 0

64

Hope this helps youuuu

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Which order pair lies on the line 2x-y= -5

Lady bird [3.3K]

Answer:

Let's solve for x.

2x−y=−5

Step 1: Add y to both sides.

2x−y+y=−5+y

2x=y−5

Step 2: Divide both sides by 2.

2x/2=y-5/2

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x=1/2y+-5/2

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3 years ago

Brian works in the lab and feels an empty cylindrical flask with 200 cubic centimeters of liquid. The flask has a radius of 3 cm

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Answer:

Step-by-step explanation:

volume of liquid in the flask= 200 cm^3

radius of the flask = 3 cm

height of the flask= 12 cm

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3 years ago

Which expression shows how 6 • 45 can be rewritten using the distributive property?

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6*45=270 is the first step you should do to get the answer you are looking for. The next thing I would do is go to all the answers to see which one comes out the same way. The first one comes to be 246, the next one is 54, the next is 270 and the last one is 220. 6*40+6*5 is the correct answer.

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3 years ago

Which of these is the standard form of the following function?<br> f(n) = 7(7 - 1)2 + 3

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3 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 $0.31\; \rm m$ .

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2 years ago