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EleoNora [17]
3 years ago
12

What is the general form of the equation of a circle with its center at (-2,1) and passing though (-4,1) ?

Mathematics
1 answer:
julsineya [31]3 years ago
4 0
Equation is :( x + 2 )² + ( y - 1 )² = r²( - 4 + 2 )² + ( 1 - 1 )² = r²4 = r², than we will plug in to a formula:( x + 2 )² + ( y - 1 )² = 4x² + 4 x + 4 + y² - 2 y +1 = 4x² + y² + 4 x - 2 y + 1 = 0
the answer is : <span>x2 + y2 + 4x − 2y + 1 = 0 </span>
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3x+1=8x-24 helpppppppp
Irina18 [472]

3x+1 =8x -24

3x -8x +1 = -24

-5x +1 = -24

-5x = -24 -1

-5x = -25

x = -25/-5 = 5

x = 5

4 0
3 years ago
Read 2 more answers
Need help with mean , median , mode question<br> + a few more..
Andreas93 [3]

Answer:

Mean:

a) 4

b) 2

c) 2

d) 2

Median:

a) 4,2

b) 5,2

c) 5,1

d) 4,1

Mode:

a) 1

b) 1

c) 1

d) 1

Step-by-step explanation:

Mean:

a) 1+4+2+1 = 8/4=4

b)1+5+2+1 = 9/4= 2.25 (round to the nearest whole number = 2)

c) 1+5+1+0 = 7/4 = 1.75 (round to the nearest whole number = 2)

d) 1+4+1+0 = 6/4 = 1.5 (round to the nearest whole number = 2)

3 0
3 years ago
If tan A = 4/3 and sin B = 45/53 and angles A and B are in Quadrant I, find the value of tan(A+B)
lbvjy [14]

Answer:

First, find tan A and tan B.

cosA=35 --> sin2A=1−925=1625 --> cosA=±45

cosA=45 because A is in Quadrant I

tanA=sinAcosA=(45)(53)=43.

sinB=513 --> cos2B=1−25169=144169 --> sinB=±1213.

sinB=1213 because B is in Quadrant I

tanB=sinBcosB=(513)(1312)=512

Apply the trig identity:

tan(A−B)=tanA−tanB1−tanA.tanB

tanA−tanB=43−512=1112

(1−tanA.tanB)=1−2036=1636=49

tan(A−B)=(1112)(94)=3316

kamina op bolte

✌ ✌ ✌ ✌

3 0
2 years ago
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Andre45 [30]
66 is the answer
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5 0
2 years ago
What is the answer to 2 3/4 ÷ 1 5/8
alex41 [277]
Exact form:
22/13

Mixed number:
1 9/13
6 0
2 years ago
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