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2 years ago

Please help me!!!!!!!!!

Mathematics

2 answers:

gizmo_the_mogwai [7]2 years ago

8 0

Answer:

396

Step-by-step explanation:

You multiply 3 times 3 times 4 times 11 to find your answer 396.

Pavel [41]2 years ago

3 0

You could split them in two different shapes either a rectangle and square or two rectangles and add up all sides of one then do the same to the other and add them both

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F(x)= x squared + 6x + 8

defon

Answer: (x+4) (x+2)

Step-by-step explanation:

x^2 + 2x + 4x +8

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2 years ago

keon ha some change in his pocket then a friend loaned him 0.25 now keon has 1.45 in his pocket which equation can be used to fi

Studentka2010 [4]

Answer:

$1.45-$0.25=x

x=$1.20

Step-by-step explanation:

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2 years ago

Determine the verticle asymptote and horizontal asymptote of x^2-25/2x^+13x+15?

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Answer:

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5 0

2 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.

case a) $x^{2}+2x+4=0$

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

case b) $x^{2}-2x+4=0$

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

case c) $x^{2}+2x-4=0$

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

case d) $x^{2}-2x-4=0$

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

the answer is

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

5 0

2 years ago

Read 2 more answers

What is 7÷10 in faction form and improper fraction form

leonid [27]

Answer:

a) 7/10

b) There is no improper fraction form for this

5 0

2 years ago