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cupoosta [38]
3 years ago
12

Which is the equation of the circle that has a diameter with endpoints located at (-2, 0) and (6, 6)?

Mathematics
1 answer:
Tpy6a [65]3 years ago
5 0
(x-2)^2 + (y-3)^2 = 25


center is 2,3
diameter is 10
rad is 10/2 = 5


Mark brainliest please
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U= V H2a + m. solve for a
fgiga [73]

Answer:

Step-by-step explanation:

For greater clarity let's write u= V H2a + m as u = vh^2a.

We can isolate a by dividing both sides of this equation by vh^2:

    u

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Alika [10]

Answer: C

Step-by-step explanation:

No matter how big the number you are dividing by you cannot go below 0.

(positive division)

so that rules out A & B.

(D) 10 - 100 = -90.

(C) 1 - 100 = -99.

C has the least value.

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4 years ago
Use the four-step process to find f'(x) and then find f'(1), f'(2), and f'(3).
Soloha48 [4]

Step 1: evaluate f(x+h) and f(x)

We have

f(x+h) = -(x+h)^2+6(x+h)-5 = -(x^2+2xh+h^2)+6x+6h-5

= -x^2-2xh-h^2+6x+6h-5

And, of course,

f(x)=-x^2+6x-5

Step 2: evaluate f(x+h)-f(x)

f(x+h)-f(x)=-x^2-2xh-h^2+6x+6h-5-(-x^2+6x-5)=-2xh-h^2+6h

Step 3: evaluate (f(x+h)-f(x))/h

\dfrac{f(x+h)-f(x)}{h}=-2x-h+6

Step 4: evaluate the limit of step 3 as h->0

f'(x) = \displaystyle \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}=-2x+6

So, we have

f'(1) = -2\cdot 1+6 = 4,\quad f'(2) = -2\cdot 2+6 = 2,\quad f'(3) = -2\cdot 3+6 = 0

5 0
3 years ago
10 of 10
lorasvet [3.4K]

Answer:

49

Step-by-step explanation:

The best way is to answer the question in reverse.

14*4

=56

56-7

=49

double check your answers too!

49+7

=56

56/4

=14

8 0
3 years ago
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