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2 years ago

Write an algebraic expression for: the sum of a squared and 6

Mathematics

1 answer:

Darina [25.2K]2 years ago

8 0

The answer is a² + 6. "Sum of a squared and 6" means you square a value and then add 6 to it.

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What is the solution to the equation -3d/a^2-2d-8 + 3/d-4 = -2/ d+2

Alex73 [517]

Answer:

d=1

Step-by-step explanation:

$\frac{-3d}{d^2-2d-8} +\frac{3}{d-4} =\frac{-2}{d+2}$

Lets factor the denominator d^2 -2d-8

d^2 - 2d - 8 = (d-4)(d+2)

$\frac{-3d}{(d-4)(d+2)} +\frac{3}{d-4} =\frac{-2}{d+2}$

Now make the denominators same

LCD: (d-4)(d+2)

$\frac{-3d}{(d-4)(d+2)} +\frac{3(d+2)}{(d-4)(d+2)} =\frac{-2(d-4)}{(d+2)(d-4)}$

Denominators are same on both sides

So equate the numerators

-3d +3(d+2) = -2(d-4)

-3d +3d +6 = -2d +8

6 = -2d + 8

subtract 8 on both sides

-2 = -2d

So d=1

3 0

2 years ago

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Random Weirdos Unite, 'm bored ._.

tresset_1 [31]

Answer:

Same here.

Step-by-step explanation:

I'm usually bored, that's why I mainly use this site.

4 0

3 years ago

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I really need help with this no sum answers or else. You will get 20 points and brain list

nordsb [41]

Answer:

if their asking for % then this is the answers

1. 41% of 900 is 369

2. 24%

3. 46%

4. 23%

5. 95%

6. 23%

7. 75%

8. 22%

9. 92%

10. 45%

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2 years ago

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A circle centered at the origin has a radius of 5 units. The terminal side of an angle, , intercepts the circle in Quadrant 3 at

rodikova [14]

Answer:

sorry

Step-by-step explanation:

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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.

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

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

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3 years ago

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