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Nezavi [6.7K]
3 years ago
13

If Anna walled for 5 miles and then turned around and went back what is her displacement?

Mathematics
1 answer:
damaskus [11]3 years ago
3 0
If she went back 5 miles than the displacement would be 0.

Just subtract the amount of miles she went back from the miles she went forward 5 - 5 = 0
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I need help please. ​
Andrei [34K]

Answer: $556.50

Step-by-step explanation: 6% of 525 is 31.5 so 525 + 31.5 = 556.50$

8 0
3 years ago
Write the linear equation in slope-intercept form -4x + 2y = 10
ella [17]
You know slope intercept form is y=mx+b so just put in that form by:
Adding 4x to other side
2y=10+4x
Then divide by 2
Y=10/2 +4/2x
Simplify
Y=5+2x
I rearranged so x is first so your answer would be: y=2x+5
3 0
3 years ago
Write the slope intercept form through (-1, -1), perp to y= 1/3x -3
amid [387]

Answer:

y = - 3x - 4

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = \frac{1}{3} x - 3 ← is in slope- intercept form

with slope m = \frac{1}{3}

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{\frac{1}{3} } = - 3 , thus

y = - 3x + c ← is the partial equation

To find c substitute (- 1, - 1) into the partial equation

- 1 = 3 + c ⇒ c = - 1 - 3 = - 4

y = - 3x - 4 ← equation of perpendicular line

6 0
3 years ago
What is:<br> X/6+4=15<br> In simplest form?
defon
X/6+4=15
subtract 4 from both sides to get rid of the addition
x/6=11
multiply by 6 from both sides to get rid of the division
x=66(simplest form)
5 0
3 years ago
X^3-3x^2-10x+24 rational zero theorem
kvasek [131]
The rational zero theorem looks at the number of sign changes in the equation's terms when x is evaluated with a positive or negative number. In your problem, if we insert a positive value for x:
1st term stays +, 2nd term stays  -, 3rd term stays -, last term stays +.
Read from left to right and determine the number of sign changes:
+ - - + [there is a change from + to -, no change from - to -, and a change from - to +] Therefore 2 sign changes. There can be two positive rational zeros, or zero rational zeros [It's always the number of sign changes, and then every other number less than the number of sign changes, and zero]

Let's now do it for a negative x:
1st term becomes -, 2nd term stays -, 3rd term becomes +, last term stays +
- - + + [there is no sign change from - to -, one change from - to +, no change from + to +] Therefore one sign change. There can be one negative rational zero, or zero rational zeros.
This is a difficult concept to type out in a screen box. Please post me a comment if you need additional clarification!
4 0
3 years ago
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