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aleksley [76]
3 years ago
12

The elevation at ground level is 0 feet. An elevator starts 90 feet below ground level. After traveling for 15 seconds, the elev

ator is 20 feet below ground level. Which statement describes the elevator's rate of change in elevation during this 15-second interval?
Mathematics
1 answer:
kramer3 years ago
4 0

Answer:

Step-by-step explanation:

i believe the answer is 7.33333333333 feet per second

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On a coordinate plane, a line goes through (negative 3, 2) and (0, 1). A point is at (3, 4). What is the equation of the line th
Novay_Z [31]

Answer:

y=3x-5

Step-by-step explanation:

step 1

Find the slope of the given line

we have

(-3,2) and (0,1)

The slope m is

m=(1-2)/(0+3)=-1/3

step 2

Find the slope of the line perpendicular to the given line

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of its slopes is -1)

so

The slope is

m=3

step 3

Find the equation of the line in point slope form

we have

m=3

point(3,4)

y-4=3(x-3) ----> equation in point slope form

Convert to slope intercept form

y=3x-9+4

y=3x-5

6 0
3 years ago
Read 2 more answers
What is the simplest form of the expression? <br> PLEASE HELP.
ser-zykov [4K]

Step-by-step explanation:

I hope it's correct... Ahhh... Now it's long enough to post

6 0
3 years ago
A ticket printer made 450 tickets in 9 minutes. Another ticket printer made tickets for 5 minutes at twice the rate of the first
OlgaM077 [116]

Answer:

500

Step-by-step explanation:

Rate at which first ticket printer works is 450/9 = 50 tickets per minute

Speed of second ticket printer is twice than that of the first.

So, speed of second printer is 100 tickets per minute.

In 5 minutes,second printer made 500 tickets

8 0
2 years ago
Help againnn. Yes, I’m kinda slow
Leni [432]

So, the right option is 1. so if my ans was helpful u can follow me.

5 0
3 years ago
compute the projection of → a onto → b and the vector component of → a orthogonal to → b . give exact answers.
Nina [5.8K]

\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})

We have been given two vectors $\vec{a}$ and $\vec{b}$, we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of$\vec{b}$onto $\vec{a}$means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

#SPJ4

3 0
1 year ago
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