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3 years ago

The answer is what I need help really bad I’m a athlete that needs help

Mathematics

1 answer:

DENIUS [597]3 years ago

5 0

Answer:

Its C. The 2nd to last one!!! (x-3)^(2)=20

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Determine the radius and center of the circle with the following equation <br> x^2+y^2-12x+24y-10=6

PolarNik [594]

Answer:

Center: (6, -12)

Radius: 4

Step-by-step explanation:

<u>Complete the squares</u>

<u /> $x^2+y^2-12x+24y-10=6\\\\x^2-12x-10+y^2+24y=6\\\\x^2-12x-10+46+y^2+24y+144=6+46+144\\\\(x^2-12x+36)+(y^2+24y+144)=196\\\\(x-6)^2+(y+12)^2=196$

Comparing with the standard form equation $(x-h)^2+(y-k)^2=r^2$ , since $r^2=196$ , then the radius of the circle is $r=14$ . Also, the center of the circle would be $(h,k)\rightarrow(6,-12)$ .

8 0

2 years ago

Solve the system of equations

34kurt

Answer:

y=17/2 and x=15/8

Step-by-step explanation:

By subsitution, we get 4x+4x+1=16, so 8x=15. Thus, x=15/8. Next, we plug x in to find y. y=4*15/8 +1, or 15/2 +1 = 17/2

5 0

3 years ago

Read 2 more answers

The legs of a right triangle are 29.25 and 13. What is the hypotenuse?

Ivan

Answer:

32.008

Step-by-step explanation:

4 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

the recipe for pumpking pie instructs you to bake the pie at 425∘F, for 15 minutes and then reduce the oven temperature to 350∘F

lakkis [162]

Answer:

ΔT = -75°F

Step-by-step explanation:

ΔT = T₁ - T₀ = 350 - 425 = -75°F

4 0

3 years ago