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2 years ago

7

(THREE CAPITAL LETTERS FOR EACH) ANSWER CORRECTLY !!!!! WILL MARK BRIANLIEST !!!!!! URGENT !!!!!!!!!!

1 answer:

Natasha2012 [34]2 years ago

4 0

FHG is congruent EFH

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Pleaseeeee helpppppppppp

Viktor [21]

Answer:

im scared for my future, I dont wanna learn about this, it looks hard

Step-by-step explanation:

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2 years ago

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What is -7 to the second power

Paha777 [63]

Answer:

+49

Step-by-step explanation:

-7 x -7 = +49 because the negatives cancel out, resulting in a positive

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2 years ago

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Find the missing length ‘x’

givi [52]

Answer:

x = 13

Step-by-step explanation:

The area (A) of a trapezium is calculated as

A = $\frac{1}{2}$ h (b₁ + b₂ )

where h is the perpendicular height and b₁, b₂ the parallel bases

Here A = 55, h = 5. b₁ = x, b₂ = 9, then

$\frac{1}{2}$ × 5 × (x + 9) = 55

2.5(x + 9) = 55 ( divide both sides by 2.5 )

x + 9 = 22 ( subtract 9 from both sides )

x = 13

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3 years ago

Write an equation for the following problem

Novosadov [1.4K]

Answer:

I believe I know the answer! Please do tell me If I am wrong but math was never my strong sute so tell me If I’m wrong please!

Step-by-step explanation:

8n-15=27 or

(8xn)-15=27

Bonus: n= 5.25 or 5 1/4

(I think)

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2 years ago

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