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musickatia [10]
2 years ago
7

Express as a difference.

Mathematics
1 answer:
ElenaW [278]2 years ago
3 0
10

b: 5+5
c: 5+5

thats thw andeer
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Help me please I NEED to pass this
Alina [70]

OPTION C is the correct answer.

Hope it helps you.

4 0
3 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
How to solve long division by using long division 711.9 divided by 8.4
harkovskaia [24]
Multiply both 711.9 and 8.4 by 10 which will remove the decimal point. 
and make the calculation much more easier
 the required answer would be 84.75

5 0
3 years ago
Can someone help me with this
mash [69]

The center of the circle is (h,k) = (-2,-3)

The radius of the circle is r = 2

The standard form of equation of the circle is

{(x + 2)}^{2}  +  {(y + 3)}^{2}  = 4

<h3>How to find the center, radius and standrad form of the circle?</h3>

The general form of equation of the circle is

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

Here, (h,k) means centre of the circle.

r means radius of the circle.

given that coordinate points of centre of circle is (-2,-3).

Hence the (h,k) = (-2,-3)

<h3>How to find the radius of the circle?</h3>

Now to find the radius of the circle

The distance from a circle's centre to its circumference is its radius.

The distance from a circle's centre (-2,-3) to its circumference (0,-3) is its radius.

using the formula, distance between the two points to obtain radius.

d =  \sqrt{(x1 - x2) {}^{2}  +  {(y1 - y2)}^{2} }  \\ r =  \sqrt{ {( - 2 - 0)}^{2} +  {( - 3 - ( - 3))}^{2}  }  \\ r =  \sqrt{ {( - 2)}^{2} +  {( - 3 + 3)}^{2}  }  \\ r =  \sqrt{ {4}^{2} + 0 }  \\ r =  \sqrt{4}  \\ r = 2

<h3>How to find the standard form of equation of the circle?</h3>

(h,k) = (-2,-3)

r = 2

subtitue the (h,k) and r values to get the standard form of equation of the circle.

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

{(x - ( - 2))}^{2}  +  {(y - ( - 3))}^{2} =  {r}^{2}

{(x + 2)}^{2}  +  {(y + 3)}^{2}  = 4

Learn more about circle, refer:

brainly.com/question/24810873

#SPJ9

5 0
1 year ago
Please help i don’t understand
SashulF [63]

Answer:

Use inverse cos to solve for x.

hope this helps

3 0
2 years ago
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