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Vesnalui [34]
3 years ago
15

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

3 0
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:

7 0
2 years ago
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Double-Angle and Half-Angle Identiies [See Attachment] Question 4
Nesterboy [21]
Please make it brainleist

8 0
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

6 0
3 years ago
Read 2 more answers
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}:

<u>For the first parenthesis:</u>

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}

<u>For the second parenthesis:</u>

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)

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

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
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