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3 years ago

On a number line what is the distance between -12 and 20 ??

Mathematics

1 answer:

KIM [24]3 years ago

3 0

Answer: Step-by-step explanation:

In order to answer this question you will need to find 2 distances and then add them to each other.

Begin at -29. How far will you have to move to the right 29 in order to get to 0?

Now how far will you have to more to get from 0 to 100?

Draw an open number line to help you.

_____________________________________________________

-29 0 100

Your first distance is 29. Your second distance is 100. The total distance is 129.

You might be interested in

What is an equation of the line that passes through the points (-8, 2) and (-4, -3)?

denpristay [2]

Answer:

y = (-6/13)x + (4/13).,

Step-by-step explanation:

the equation of the line is:

y = mx + b, where "m" is the slope and "b" gives the y-intercept

m = (y2 - y1)/(x2 - x1)

m = (-2 - 4)/(5 - (-8))

m = -6/13

y = (-6/13)x + b

the line passes through the point (-8,4) means that for x = -8, y = 4

4 = (-6/13)(-8) + b

b = 4 - (-6/13)(-8)

b = 4/13

the equation of the line that passes through the points (-8,4) and (5,-2) is:

y = (-6/13)x + (4/13).

5 0

3 years ago

Why does a calculator return an error when trying to find inverse sine of 1.055?

anyanavicka [17]

Sin(x) is a function bounded between -1 and 1. A value greater than 1 or smaller than -1 (-1.1, for instance) does not correspond to any real angle.

It then returns error

6 0

3 years ago

Read 2 more answers

A street light is at the top of a 25 ft pole. A 4 ft tall girl walks along a straight path away from the pole with a speed of 6

Burka [1]

Answer:

Tip of the shadow of the girl is moving with a rate of 7.14 feet per sec.

Step-by-step explanation:

Given : In the figure attached, Length of girl EC = 4 ft

Length of street light AB = 25 ft

Girl is moving away from the light with a speed = 6 ft per sec.

To Find : Rate ( $\frac{dw}{dt}$ ) of the tip (D) of the girl's shadow (BD) moving away from th

light.

Solution : Let the distance of the girl from the street light is = x feet

Length of the shadow CD is = y feet

Therefore, $\frac{dx}{dt}=6$ feet per sec. [Given]

In the figure attached, ΔAFE and ΔADE are similar.

By the property of similar triangles,

$\frac{x}{21}=\frac{x+y}{25}$

25x = 21(x + y)

25x = 21x + 21y

25x - 21x = 21y

4x = 21y

y = $\frac{4x}{21}$

Now we take the derivative on both the sides,

$\frac{dy}{dt}=\frac{4}{21}\times \frac{dx}{dt}$

= $\frac{4}{21}\times 6$

= $\frac{8}{7}$

≈ 1.14 ft per sec.

Since w = x + y

Therefore, $\frac{dw}{dt}= \frac{dx}{dt}+\frac{dy}{dt}$

$\frac{dw}{dt}=6+1.14$

= 7.14 ft per sec.

Therefore, tip of the shadow of the girl is moving with a rate of 7.14 feet per sec.

3 0

3 years ago

I need help, I don't know how to do it.

Sindrei [870]

Answer:

Step-by-step explanation:

to solve this problem we can use the Pythagorean theorem

UT and TL are the legs, while LU is the hypotenuse

We have to find LU so we can proceed like this

x^2 + (x+1)^2 = LU^2

x^2 + x^2 + 1 + 2x = LU^2

2x^2 + 2x + 1 = LU^2

LU = +/- $\sqrt{2x^2+2x+1$

we have to take only the positive value because a length can’t be negative.

2x^2 + 2x + 1 is positive for every value of x, so the final answer is

$\sqrt{2x^2+2x+1}$

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2 years ago

Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz help me

Vaselesa [24]

Show another picture i cant see you

5 0

3 years ago