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Nataly [62]
2 years ago
8

Find the slope of the line. Explain or show your reasoning.

Mathematics
1 answer:
bekas [8.4K]2 years ago
6 0
What are the points?
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I really need help please !!!
lisov135 [29]

Answer:

15

Step-by-step explanation:

(whole secant) x (external part) = (tangent)^2

(5+x) * 5 = 10^2

Distribute

25 +5x = 100

Subtract 25 from each side

25-25 +5x = 100-25

5x= 75

Divide by 5

5x/5 = 75/5

x = 15

8 0
2 years ago
What is the square root of 12 divided by 2 plus 16?
slavikrds [6]

Answer:

17.73205081

Step-by-step explanation:

Didn't use a calculator :) Please mark me as brainliest. Thanks!

4 0
3 years ago
Read 2 more answers
A train averages a speed of 90 miles per hour across the plains and 37.5 miles per hour through the mountains. If a trip of 300
densk [106]
Let p and m represent the numbers of miles in the plains and mountains, respectively.
.. p + m = 300 . . . . . . . the whole trip was 300 miles

time = distance/speed, so the trip time can be written as
.. p/90 + m/37.5 = 3 48/60
.. 5p + 12m = 1710 . . . . . . . . . . multiply by 450 to put in standard form

You can solve these two equations by any of several means. The variable p can be eliminated by subtracting 5 times the first from the second.
.. (5p +12m) -5(p +m) = 1710 -5*300
.. 7m = 210
.. m = 30

30 miles of the trip was through the mountains.
5 0
3 years ago
Read 2 more answers
An item is regularly priced at $15. It is on sale for 60% off the regular price. What is the sale price?​
olya-2409 [2.1K]

Answer:

$15 - 60% = $6.00

https://www.percent-off.com/_30_%25+off_39.98_

(I used this website because I lost my calutor, plus I use it to help my mom shop)

6 0
3 years ago
The slope of the tangent line to the curve x^3y+y^2-x^2=5 at the point (2,1) is
zheka24 [161]

Answer:

-4/5

Step-by-step explanation:

To find the slope of the tangent to the equation at any point we must differentiate the equation.

x^3y+y^2-x^2=5

3x^2y+x^3y'+2yy'-2x=0

Gather terms with y' on one side and terms without on opposing side.

x^3y'+2yy'=2x-3x^2y

Factor left side

y'(x^3+2y)=2x-3x^2y

Divide both sides by (x^3+2y)

y'=(2x-3x^2y)/(x^3+2y)

y' is the slope any tangent to the given equation at point (x,y).

Plug in (2,1):

y'=(2(2)-3(2)^2(1))/((2)^3+2(1))

Simplify:

y'=(4-12)/(8+2)

y'=-8/10

y'=-4/5

5 0
2 years ago
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