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2 years ago

Line p intersect lines a and b

Mathematics

1 answer:

ser-zykov [4K]2 years ago

7 0

solution are on the pic

hope it's helpful for you

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Help me with this!!!!

sdas [7]

Answer:

IJ = 34

Step-by-step explanation:

Since HK = JK, then IK is a perpendicular bisector and Δ is isosceles, then

IJ = HI , that is

2s = s + 17 ( subtract s from both sides )

s = 17

Then

IJ = 2s = 2 × 17 = 34

7 0

2 years ago

Read 2 more answers

Answer. it fast please

mojhsa [17]

Answer:

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{15}$

Step-by-step explanation:

Given

$(\frac{6}{10})^3 * (\frac{5}{9})^2$

Required

Solve:

$(\frac{6}{10})^3 * (\frac{5}{9})^2$

Simplify 6/10

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = (\frac{3}{5})^3 * (\frac{5}{9})^2$

Express 9 as 3^2

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = (\frac{3}{5})^3 * (\frac{5}{3^2})^2$

Apply law of indices

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3}{5^3}* \frac{5^2}{(3^2)^2}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3}{5^3}* \frac{5^2}{3^4}$

Express as a sing;e fraction

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3*5^2}{5^3*3^4}$

Apply law of indices:

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5^{3-2}*3^{4-3}}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5^1*3^1}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5*3}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{15}$

5 0

2 years ago

0.5 (2x + 2) = -4<br><br> Solve for x.

Ivanshal [37]

0.5(2x+2)=-4

Multiply by 2 on both sides

2x+2=-8

Subtract 2 on both sides

2x=-10

Divide by 2 on both sides

x=-5

3 0

2 years ago

Read 2 more answers

Are their always the same number of prime numbers between 2 consecutive multiples of 10

Harlamova29_29 [7]

Yes there are. Its just the way they work out!

3 0

3 years ago

Find <br><img src="https://tex.z-dn.net/?f=%28%20%7B4x%7D%5E%7B2%7D%20%20%2B%203x%29%20%2B%20%28%20%7B7x%7D%5E%7B2%7D%20%20-%205

marishachu [46]

Answer:

$11x^2-2x-6$

Step-by-step explanation:

$(4x^2+3x)+(7x^2-5x-6)= \\\\(4x^2+7x^2)+(3x-5x)-6= \\\\11x^2-2x-6$

Hope this helps!

4 0

3 years ago