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3 years ago

9

Find the perimeter. Simplify your answer. 85-8 9s-2 9s-2 99-10

1 answer:

stich3 [128]3 years ago

8 0

Answer:

35s-22

Step-by-step explanation:

Perimeter is found by adding the length of all sides.

(8s-8)+(9s-10)+(9s-2)+(9s-2)

35s-22

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Write the name of the period that has the digits 913

tankabanditka [31]

Thousands grou
the thousands group. 913,256 256 being the ones group 913 being the thousands group. the name of the period with 913, is the thousands group p

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Someone please help me I need help

pochemuha

Find how many mm are in 25m (by multiplying 25 by 1000) and convert that answer to km.

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3 years ago

Solve for all values of x by factoring.<br> x^2 + 4x – 30 = -3x

vichka [17]

Answer:

${x}^{2} + 4x - 30 = - 3x \\{x}^{2} + 7x - 30 = 0 \\ {x}^{2} - 3x + 10x - 30 = 0 \\ x(x - 3) + 10(x - 3) = 0 \\ (x - 3)(x + 10) = 0 \\ \boxed{x = 3} \\ \boxed{x = - 10}$

<h3><em><u>x=3 or x=-10</u></em> is the right answer.</h3>

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3 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

<u>case b)</u> $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

<u>case c)</u> $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

<u>case d)</u> $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

$x^{2}-2x-4=0$

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3 years ago

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stiv31 [10]

Answer:

A≈110.11

using formula 1/4 $\sqrt5(5+2\sqrt5)a2$

Step-by-step explanation:

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3 years ago