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2 years ago

Which is the closest to the area that is painted blue? A 308 units B. 116 units C. 463 units? D 124 units

Mathematics

1 answer:

Charra [1.4K]2 years ago

6 0

Step-by-step explanation:

first of all, all answer options are basically wrong, because an area is always measured in units² (and not just in units).

the blue colored area is the area of the whole square minus the area of the circle.

the side length of the square is 2 times the radius of the circle (the diameter) : 2×12 = 24 units.

so, the area of the square is

24² = 576 units²

the are of the circle is

pi×r² = pi×12² = 144pi units²

so, the blue area is

576 - 144pi = 123.6106579... ≈ 124 units²

therefore, D is the right answer.

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A town has a population of 12000 and grows at 4% every year. To the nearest tenth of a year, how long will it be until the popul

ch4aika [34]

Answer:

11.0 years

Step-by-step explanation:

The exponential function describing the population growth can be written ...

population = (initial population)×(1 +growth rate)^years

If t represents the number of years, we can fill in the values to get ...

18500 = 12000×1.04^t

Dividing by 12000 gives ...

18500/12000 = 1.04^t

Taking logarithms, we have ...

log(18500/12000) = t×log(1.04)

t = log(185/120)/log(1.04) ≈ 11.037 ≈ 11.0

It will be 11.0 years until the population reaches 18,500.

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3 years ago

What is the slope of the line?

Leya [2.2K]

Answer:

The slope of the line is -3

Step-by-step explanation:

I chose the points (-2,3) and (0,-3)

The slope formula: m= $\frac{y_2-y_1}{x_2-x_1}$

$\frac{-3-3}{0-(-2)}$ ==> $\frac{-3-3}{0+2}$ ==> $\frac{-6}{2}$ = $-3$

Our final answer is -3.

Hope that helps!

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2 years ago

the ratio of adults to children attending a new exhibit at the museum was found to be 8:5. based on this ratio if 390 people att

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3 years ago

5.5-x=-4.5-x what would that equale

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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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3 years ago