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mel-nik [20]
3 years ago
5

(x + 3)2 + (y – 5)2 = 1 What would be the center of this?

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
8 0

Answer:

(- 3, 5 )

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

(x + 3)² + (y - 5)² = 1 ← in standard form

with centre = (h, k) = (- 3, 5 )

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3c-d=T<br> Could you<br> solve for c
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- - Let's solve for C.

Starting off with our equation:
3c+d=T

Transform:
3c=T+D

Solve for that:
\frac{3c}{3} = \frac{T+D}{3}


Solve for that for the last time and your answer is \frac{T+D}{3} =C



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15x^2 + 41x + 28 <br><br>STEP BY STEP ANSWER PLSSS​
maria [59]

Answer:

(3x+4)(5x+7)

Step-by-step explanation:

15x^2 +41x+28

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^2 +ax+bx+28. To find a and b, set up a system to be solved.

a+b=41

ab=15×28=420

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 420.

1,420

2,210

3,140

4,105

5,84

6,70

7,60

10,42

12,35

14,30

15,28

20,21

Calculate the sum for each pair.

1+420=421

2+210=212

3+140=143

4+105=109

5+84=89

6+70=76

7+60=67

10+42=52

12+35=47

14+30=44

15+28=43

20+21=41

The solution is the pair that gives sum 41.

a=20

b=21

Rewrite 15x^2 +41x+28 as (15x^2 +20x)+(21x+28).

(15x^2 +20x)+(21x+28)

Factor out 5x in the first and 7 in the second group.

5x(3x+4)+7(3x+4)

Factor out common term 3x+4 by using distributive property.

(3x+4)(5x+7)

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