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nadezda [96]
3 years ago
12

Write the standard form of the equation of the circle with its center at (-5,0) and a radius of 8 what is the equation of the ci

rcle in standard form
Mathematics
1 answer:
Leto [7]3 years ago
7 0

Answer:

\large\boxed{(x+5)^2+y^2=64}

Step-by-step explanation:

The standard form of an equation of a circle:

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

(h, k) - center

r - radius

We have the center (-5, 0) and the radius r = 8. Substitute:

(x-(-5))^2+(y-0)^2=8^2\\\\(x+5)^2+y^2=64

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Which of the following equations have complex roots?
mestny [16]

Answer:

A

Step-by-step explanation:

Complex roots of quadratic functions occur when the <u>discriminant is negative</u>.

<u>Discriminant</u>

b^2-4ac\quad\textsf{when}\:\:ax^2+bx+c=0

Evaluate the discriminant of each of the given equations.

\textsf{A.} \quad 3x^2+2=0

\implies a=3, \quad b=0, \quad c=2

\implies b^2-4ac=0^2-4(3)(2)=-24

As -24 < 0 the equation will have complex roots.

\textsf{B.} \quad 2x^2+1=7x

\implies 2x^2-7x+1=0

\implies a=2, \quad b=-7, \quad c=1

\implies b^2-4ac= (-7)^2-4(2)(1)=41

As 41 > 0 the equation does not have complex roots.

\textsf{C.} \quad 3x^2-1=6x

\implies 3x^2-6x-1=0

\implies a=3, \quad b=-6, \quad c=-1

\implies b^2-4ac=(-6)^2-4(3)(-1)=48

As 48 > 0 the equation does not have complex roots.

\textsf{D.} \quad 2x^2-1=5x

\implies 2x^2-5x-1=0

\implies a=2, \quad b=-5, \quad c=-1

\implies b^2-4ac=(-5)^2-4(2)(-1)=33

As 33 > 0 the equation does not have complex roots.

Learn more about discriminants here:

brainly.com/question/27444516

brainly.com/question/27869538

Learn more about complex roots here:

brainly.com/question/26344541

6 0
2 years ago
Why is there always more than one parallelogram with the same area and base​
BaLLatris [955]

Answer:

If every line parallel to two lines intersects both regions in line segments of equal length, then the two regions have equal areas. In the case of your problem, every line parallel to the bases of the two parallelograms will intersect them in lines segments, each with a width of ℓ.

6 0
2 years ago
Can you help with part c I have a and b answered
Vlad [161]
I hope this helps you

8 0
3 years ago
Where do I find my work I did with a tutor?
OlgaM077 [116]
It should come up as a notification. Check your messages or press “tutor” maybe even try reloading the page.
8 0
2 years ago
Please help!<br><br> If sin(xº)=cos(yº) find k if x=2k+3 and y=6k+7
USPshnik [31]

answer

10

step-by-step explanation

the equation given is sin(x) = cos(y) with x = 2k + 3 and y = 6k + 7

substitute in 2k+ 3 for x in sin(x) and substitute in 6k + 7 for y in cos(y)

sin(x) = cos(y)

sin(2k + 3) = cos(6k + 7)

we know that sin(x) = cos(90 -x)

sin(2k + 3)

= cos(90 - (2k + 3) )

= cos(90 - 2k - 3)

= cos(87 - 2k)

substitute this into the equation sin(2k + 3) = cos(6k + 7)

sin(2k + 3) = cos(6k + 7)

cos(87 - 2k) = cos(6k + 7)

87 - 2k = 6k +7

80 = 8k

k = 10

7 0
3 years ago
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