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2 years ago

HELPPPPPPPPP!

Mathematics

2 answers:

bekas [8.4K]2 years ago

4 0

Answer:

30%

Step-by-step explanation:

So she spun the spinner 30 times so we divide the number of times she got solid by the 30 to get the answer in a percentage :)

9/30 = 0.3

Now we multiply the 0.3 by 100 to get it as a percent!

0.3 x 100 = 30%

Now you know how to find the probability of an object!

Have a great day!!

Please rate and mark brainliest!!

emmasim [6.3K]2 years ago

3 0

<h3>Answer: 3/10</h3>

Explanation:

Count the tally marks for the "solid" in the middle row. The first portion is a group of 5 as shown by the slash mark. The second portion of that row is one short of five, so it's another 4 tallies. Overall, we have 5+4 = 9 tallies for the "solid".

This is out of 30 spins total. Therefore, we arrive at 9/30 = (3*3)/(3*10) = 3/10

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PLEASE HELP ME IM LOST! WILL GIVE BRAINLIEST!!!! ABC has vertices A(0,6) B(4,6) C(1,3). Sketch a graph of ABC and use it to find

Paul [167]

The orthocenter is the point where the three altitudes meet.
sketch the graph and you will see that AB is a horizontal line, the altitude is a vertical line through the point (1,3), so the equation of this altitude is x=1

next, find another altitude. I'll use the altitude of BC.
the slope of BC is (6-3)/(4-1)=1, so the slope of the altitude, which is perpendicular to BC going through the point A (0,6), is -1, the equation of the altitude of BC is y=-x+6
the system of equation : x=1
y=-x+6
has a solution (1, 5)
the solution is where the two lines meet, the meeting point is the orthocenter.

Double check by find the equation for other altitude:
slope of AC: (3-6)/(1-0)=-3
slope of altitude of AC: 1/3
equation of altitude of AC: y=(1/3)x+b
the altitude of AC goes through point B (4,6), so we can find out b by plug x=4, y=6 in the equation: 6=(1/3)*4+b, b=14/3
y=(1/3)x+14/3
Is (1,5) also a solution to this equation? Plug x=1 in the equation, we get y=5, so yes, (1,5) is a point on the third altitude.

7 0

3 years ago

Solveeee please :(<br> 6/25 × 3/4 ÷ 16/5

yarga [219]

Answer:

6/25 x 3/4 = 18/100 % 16/5 = 18/100 x 5/16 = 90/1600 or 0.05625

Can also be simplified to 9/160

3 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.