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3 years ago

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THE algebra tiles represent the perfect square trionmaial x^2+10x+c

1 answer:

aleksandr82 [10.1K]3 years ago

3 0

$( { \frac{b}{2} )}^{2} = c \\ ( { \frac{10}{2}) }^{2} \\ ( {5)}^{2} \\ 25 = c$

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Lola takes the train from Paris to Nice. The distance between the two cities is about 900,000 meters. If the train travels at a

Elanso [62]

Nice??? there is a city called nice? anyways, your answer is 5000. hope this helps :)

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3 years ago

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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>

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

5 0

3 years ago

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Solve this n get pts :)

Simora [160]

D+100+40=180
d=180-100-40=40 (sum of all angles)
e= 100+40= 140 (exterior angle)
180-140=40
C+50+40=180
C=180-50-40= 90 (right angle)
A=50
B=180-90-50= 40

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2 years ago

What is the 4th square number

Oksanka [162]

Answer:

16 is the 4th square number.

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What is the base of a triangle that has a height of 6 centimeters and an area of 18 centimeters?

klasskru [66]

Answer:

B = 6 centimeters

Step-by-step explanation:

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