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3 years ago

What is the center of (x + 1)2 + (y - 3)2 = 9?

1 answer:

katrin2010 [14]3 years ago

5 0

The equation of a circle with center at (h,k) is
(x-h)^2+(y-k)^2=radius^2

given
(x-(-1))^2+(y-3)^2=9
center is (-1,3)

first option

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Area of triangle1/2b*h Now you can do

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3 years ago

1,323 miles 443 miles and 409 Round to nearest ten

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470

Step-by-step explanation:

1323 rounds to 1320

443 rounds to 440

409 rounds to 410

1320 - (440 + 410) =

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A triangle is graphed in the coordinate plane. The vertices of the triangle have coordinates (–5, 2), (7, 2), and (7, –3).

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8 0

3 years ago

Read 2 more answers

AB = 44 + 3xBC = 33 + yCD = 64 − xAD = 48 − 2yQuadrilateral ABCD is a parallelogram if both pairs of opposite sides are congruen

Otrada [13]

The lengths (in centimeters) of the opposite side pairs are 59 cm , 38 cm. Option C) is the correct answer.

Step-by-step explanation:

step 1 :

Given that,

Quadrilateral ABCD is a parallelogram if both pairs of opposite sides are congruent.

step 2 :

The opposite sides are AB and CD respectively.

The another opposite sides are BC and AD respectively.

step 3 :

If both pairs of opposite sides are congruent, then

AB = CD

44+3x = 64-x

3x+x = 64-44

4x = 20

x = 5

step 4 :

BC = AD

33+y = 48-2y

y+2y = 48-33

3y = 15

y = 5

step 5 :

Subsitute x=5 and y=5 in any of the given sides,

CD = 64-x = 64-5 = 59

∴ CD = 59 cm

BC = 33+y = 33+5 = 38

∴ BC = 38 cm

The lengths of the opposite side pairs are 59 cm , 38 cm.

3 0

3 years ago

a square painting has an area of 81x^2-90x-25. A second square painting has an area of 25x^2+30x+9. What is an expression that r

galina1969 [7]

Answer:

The answer in the procedure

Step-by-step explanation:

Let

A1 ------> the area of the first square painting

A2 ----> the area of the second square painting

D -----> the difference of the areas

we have

$A1=81x^{2}-90x-25$

$A2=25x^{2}+30x+9$

case 1) The area of the second square painting is greater than the area of the first square painting

The difference of the area of the paintings is equal to subtract the area of the first square painting from the area of the second square painting

D=A2-A1

$D=(25x^{2}+30x+9)-(81x^{2}-90x-25)$

$D=(-56x^{2}+120x+34)$

case 2) The area of the first square painting is greater than the area of the second square painting

The difference of the area of the paintings is equal to subtract the area of the second square painting from the area of the first square painting

D=A1-A2

$D=(81x^{2}-90x-25)-(25x^{2}+30x+9)$

$D=(56x^{2}-120x-34)$

4 0

4 years ago