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3 years ago

Write the standard form of the equation of the circle with center (8,−1) that passes through the point (6,7)

Mathematics

1 answer:

marshall27 [118]3 years ago

5 0

Answer:

(x - 8)^2 + (y + 1)^2 = 68

Step-by-step explanation:

The standard form of the equation of the circle with center (8,−1) is :

(x - 8)^2 + (y + 1)^2 = R^2

If the circle passes through the point (6,7) that means that the point (6,7) is a solution of the equation and we can replace (x,y) with (6,7) to find R.

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kenny6666 [7]

Answer:

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Step-by-step explanation:

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2 years ago

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Matthew is four years younger than his brother Kevin,when Matthew was 7,how old was his brother

cestrela7 [59]

Answer:

Kevin was eleven years old.

Step-by-step explanation:

7+4=11, since the question says Matthew is younger by four years.

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3 years ago

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Complete the equation of this line in slope-intercept form y= _x-1

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Answer:

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3 years ago

3/4 + (1/3 ÷ 1/6) - (-1/2) =?

stepladder [879]

Answer:

[See Below]

Step-by-step explanation:

✦ Turn fractions into decimals:

$\frac{3}{4} =0.75$

$-\frac{1}{2} =-0.5$

(We will do something different for $\frac{1}{3}$ and $\frac{1}{6}$ .)

✦ Divide:

$\frac{1}{3}$ ÷ $\frac{1}{6}=2$

(Since the denominators are multiplies of each other we can divide them without having to change them into decimals.)

✦ Simplified Equation:

$0.75+2-(-0.5)$

✦ Add:

$0.75+2=2.75$

✦ Subtract:

$2.75-(-0.5)=3.25$

(Subtraction rule is if there is 2 negatives next to each other in a problem it turns into a positive.)

✦ So your answer would be:

$\frac{13}{4}$ (Exact Form)

$3~ \frac{1}{4}$ (Mixed Number Form)

$3.25$ (Decimal Form)

~<em>Hope this helps Mate. If you need anything feel free to message me. </em>

8 0

3 years ago

Write as a product 27a^3−(a−b)^3

jarptica [38.1K]

Answer:

(2a +b)·(13a^2 -5ab +b^2)

Step-by-step explanation:

The factorization of the difference of cubes is a standard form:

(p -q)^3 = (p -q)(p^2 +pq +q^2)

Here, you have ...

p = 3a
q = (a-b)

so the factorization is ...

(3a -(a -b))·((3a)^2 +(3a)(a -b) +(a -b)^2) . . . . substitute for p and q

= (2a +b)·(9a^2 +3a^2 -3ab +a^2 -2ab +b^2) . . . . simplify a bit

= (2a +b)·(13a^2 -5ab +b^2) . . . . . . collect terms

6 0

3 years ago