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3 years ago

7

L is between points O and P. OP is 25 and LP is 17. What is OL? (Write as a number only)

1 answer:

allsm [11]3 years ago

3 0

Answer:

8

Step-by-step explanation:

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The answer is n= 35/2

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What property is illustrated by 5*6=6*5

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The commutative property

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Simplify ^75 select the correct answer

Ilia_Sergeevich [38]

Answer:

to simplify the square root of 75, first I recognize that the perfect square 25 goes into it. So 75 = 25 x 3. Therefore the square root of 75 equals the square root of 25 times the square root of 3. And that means just 5 times the square root of 3.

Step-by-step explanation:

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3 years ago

Distance between parallel lines y=3x+10 and y=3x-20

Alecsey [184]

1. Take an arbitrary point that lies on the first line y=3x+10. Let x=0, then y=10 and point has coordinates (0,10).

2. Use formula $d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$ to find the distance from point $(x_0,y_0)$ to the line Ax+By+C=0.

The second line has equation y=3x-20, that is 3x-y-20=0. By the previous formula the distance from the point (0,10) to the line 3x-y-20=0 is:

$d=\dfrac{|3\cdot 0-10-20|}{\sqrt{3^2+(-1)^2}}=\dfrac{30}{\sqrt{10}}=3\sqrt{10}$ .

3. Since lines y=3x+10 and y=3x-20 are parallel, then the distance between these lines are the same as the distance from an arbitrary point from the first line to the second line.

Answer: $d=3\sqrt{10}$ .

4 0

3 years ago

What's the answer for this question? (The numbers after the letters are indexes btw) 27a9 x 18b5 x 4c2 Over 18a4 x 12b2 x 2c

zysi [14]

Answer:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

Step-by-step explanation:

Given

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Required

Simplify

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Cancel out 18

$\frac{27a^9 * b^5 * 4c^2 }{a^4 * 12b^2 * 2c}$

Divide 4 and 2

$\frac{27a^9 * b^5 * 2c^2 }{a^4 * 12b^2 *c}$

Divide 27 and 12 by 3

$\frac{9a^9 * b^5 * 2c^2 }{a^4 * 4b^2 *c}$

Apply law of indices

$\frac{9a^{9-4} * b^{5-2} * 2c^{2-1} }{4}$

$\frac{9a^5 * b^3 * 2c }{4}$

Divide 2 and 4

$\frac{9a^5 * b^3 * c}{2}$

$\frac{9a^5b^3c}{2}$

Rewrite as:

$\frac{9}{2}a^5b^3c$

Hence:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

4 0

3 years ago