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3 years ago

Don't know that answer

Mathematics

2 answers:

labwork [276]3 years ago

8 0

The answer is C or Area.

alexdok [17]3 years ago

3 0

The answer is C. Area

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Find the area of the shape below

Westkost [7]

Answer:

46.875

Step-by-step explanation:

A= l x w x h so you multiple all three numbers together and get your answer

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3 years ago

Use the Distributive Property of Multiplication over Addition to find each missing number.

Svet_ta [14]

Answer:

(3*9)+(3*6)

Step-by-step explanation:

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3 years ago

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Let f(x)=x2+15x+56

Ivan

Answer:

-8, -7

Step-by-step explanation:

f(x)=x^2+15x+56

0 =x^2+15x+56

What two numbers multiply together to make 56 and add together to 15

7*8 = 56

7+8=15

0 = (x+8) (x+7)

Then use the zero product property

x+8 =0 x+7 =0

x+8-8 =0-8 x+7-7 =0-7

x=-8 x=-7

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3 years ago

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A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :

For the first parenthesis:

$x^{2}+6x+b^{2}$

We can rewrite this as:

$x^{2}+2(3)x+b^{2}$

Hence in this case $b=3$ and $b^{2}=9$ :

$x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}$

For the second parenthesis:

$y^{2}-8y+b^{2}$

We can rewrite this as:

$y^{2}-2(4)y+b^{2}$

Hence in this case $b=-3$ and $b^{2}=9$ :

$y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}$

Then, equation (5) is rewritten as follows:

$(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16$ (6)

Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.

Rearranging:

$(x-3)^{2}+(y-4)^{2}=100$ (7)

At this point we have the circle equation in the center radius form $(x-h)^{2} +(y-k)^{2}=r^{2}$

Hence:

$h=-3$

$k=4$

$r=\sqrt{100}=10$

8 0

3 years ago