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Murljashka [212]
3 years ago
5

Find an equation of the circle having the given center and radius. Center ( - 1, 1), radius 7

Mathematics
2 answers:
Alinara [238K]3 years ago
8 0

Answer:

(x+1)^2+(y-1)^2=49

Step-by-step explanation:

The standard equation of a circle is

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

where the center is (h,k) and r is the radius

so h=-1 and k=1 in this case

so then (x- -1)^2+(y-1)^2=7^2

(x+1)^2+(y-1)^2=49

suter [353]3 years ago
7 0

Answer:

(x+1)^2 + (y+1)^2 = 49

Step-by-step explanation:

We can write the equation of a circle as

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

where (h,k) is the center and r is the radius

(x- -1)^2 + (y- -1)^2 = 7^2

Simplifying

(x+1)^2 + (y+1)^2 = 49

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Turn 2 1/2 into a percentage
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Answer:

250%

Step-by-step explanation:

(2 1/2) × 100%

= 250%

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2 years ago
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Given line BD is congruent to BC, find the measure of angle B
Degger [83]
If BD is congruent to BC, that means that the sides are equal, so their angles are too.
6x-9 = 3x+24
3x = 33
x = 33/3
x = 11
Angle BCD:
6×11-9 = 66-9 =57°
Angle BDC:
3×11+24 = 33+24 = 57°
Angle B:
x = 180° - 2×57°
x = 66°
5 0
3 years ago
The first five terms of a sequence are
Elena L [17]

Answer:

(a) 7 + 3( n - 1 )

(b) 1006

Step-by-step explanation:

ARITHMETIC SEQUENCE.

The number of term of an Arithmetic progression has the formular :

nth term = a + ( n - 1 ) d

a = 7

d = 10 - 7 = 3

nth term = 7 + ( n - 1 ) 3

= 7 + 3n - 3

= 7 + 3( n - 1 )

Therefore,

the nth term of the sequence

= 7 + 3( n - 1 )

(b) For the 1000th term

= 7 + 3 ( 1000-1 )

= 7 + ( 999 )

7 + 999 = 1006

Therefore,

the 1000th term = 1006

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2 years ago
What is the quadratic function f(x)=x^2+6x-2 In vertex form?
Yuliya22 [10]

Answer: Option D

f(x)=(x+3)^2 -k

Step-by-step explanation:

For a quadratic function of the form

ax ^ 2 + bx + c

The x coordinate of the vertice is:

x =-\frac{b}{2a}

In this case the function is:

f(x)=x^2+6x-2\\\\

So

a=1\\b=6\\c=-2

The x coordinate of the vertice is:

x=-\frac{6}{2*1}\\\\x=-3

The y coordinate of the vertice is:

f(-3) = (-3)^2 +6(-3) -2\\\\f(-3)=-11

The vertice is: (-3, -11)

The form e vertice for a quadratic equation is:

f(x)=(x-h)^2 +k

Where

the x coordinate of the vertice is h and the y coordinate of the vertice is k.

Then h=-3 and k =-11

Finally the equation f(x)=x^2+6x-2\\\\ in vertex form is:

f(x)=(x+3)^2 -k

6 0
3 years ago
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The factored form of the equation is (x + 4)(x – 6) = 0.
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