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3 years ago

How do I solve this?

Mathematics

1 answer:

harina [27]3 years ago

6 0

Radical probably it could work with a protractor

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Please show me step by step so I can learn to explain it to my son.

aleksklad [387]

These are the steps:
1. Find the area of the trapezium {Whole figure).
2. FInd the area of the rectangle (unshaded).
3. Area of the shaded = Area of trapezium - Area of the rectangle.

Step 1: Find the area of the trapezium:

Formula : Area of trapezium = 1/2 (a + b)h

Area = 1/2 ( 25 + 15) (12) = 240 yd²

Step 2 : Find the area of the rectangle:

Formula : Area = Length x Width

Area = 12 x 3 = 36 yd²

Step 3: Find the shaded region:

240 - 36 = 204 yd²

Answer: 204 yd²

6 0

3 years ago

it was recently estimated that domestic vehicles out numberforegin vehicles by about six to five. if there are 2673 vehicles in

klio [65]

The question is an example of a ratio problem. This implies that

$\begin{gathered} \text{Domestic vehicles ratio=6} \\ \text{Foreign vehicles ratio=5} \\ Sum\text{ of vehicles ratio =5+6=11} \end{gathered}$

Therefore, le the number of domestic vehicles be x. We can say that;

$\begin{gathered} \frac{6}{11}=\frac{x}{2673} \\ 11x=16038 \\ x=\frac{16038}{11} \\ x=1458 \end{gathered}$

Answer: 1458

7 0

1 year ago

Which of the following is a quantity that could be represented by the number

Komok [63]

Answer:

2.404 x 10^6?. A. the population of Iowa. B. the diameter (in meters) of an atom. C. the thickness (in meters) of a sheet of printer paper. D. the amount of money (in US dollars) in the average checking account. .

When 2.404 x 10^6 is expressed in another form, the number is equal to 2,404,000. The population of a place would possibly be around this number. It cannot be the diameter of an atom since the number is too large not the thickness of a paper. It is not a common case also that the money in a bank account is around this number. Answer is A.

6 0

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.