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3 years ago

Which is NOT true? A. 16 + 3 = 11 + 8 B. 16 = 16 C. 16 - 4 = 7 + 6 D. 16 = 19 - 3

Mathematics

2 answers:

Yuliya22 [10]3 years ago

6 0

Answer:

Step-by-step explanation:

16+3:19

11+8:19

19=19

vova2212 [387]3 years ago

6 0

Answer:

Step-by-step explanation:

16-4=12

7+6=11

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The length of ZX is 2 units. What is the perimeter of triangle XYZ?

dezoksy [38]

Keywords

triangle,perimeter,distance, length, side, points

we know that

The perimeter of a triangle is the sum of the three length side

To find the length side calculate the distance between two points

The formula to calculate the distance between to points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step 1

Find the distance ZY

$Z(1,0)\\Y(3,1)$

substitute the values

$d=\sqrt{(1-0)^{2}+(3-1)^{2}}$

$dZY=\sqrt{5}\ units$

Step 2

Find the distance XY

$X(-1,4)\\Y(3,1)$

substitute the values

$d=\sqrt{(1-4)^{2}+(3+1)^{2}}$

$dXY=\sqrt{25}=5\ units$

Step 3

Find the perimeter of the triangle

$P=ZX+ZY+XY$

we have

$dZX=2\sqrt{5}\ units$

$dZY=\sqrt{5}\ units$

$dXY=5\ units$

Substitute

$P=(2\sqrt{5}+\sqrt{5}+5)\ units$

therefore

the answer is

$P=(3\sqrt{5}+5)\ units$

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Answer:

The answer is stratified.

Step-by-step explanation:

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3 years ago

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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.