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3 years ago

6

Someone please help this is due in an hour

1 answer:

NISA [10]3 years ago

3 0

Answer:

$65$

ther you go

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Solve each equation by taking square roots<br> 9v^2=25

lisov135 [29]

<span>9v^2=25
v^2 = 25/9
v^2 = (5/3)^2
v = + 5/3 and v = -5/3

hope it helps</span>

3 0

3 years ago

Please help! find the value of x

Serga [27]

I needed points sorry

4 0

3 years ago

Read 2 more answers

A circle has the equation x2 + (y−7)2 = 121. What is the length of the diameter of the circle?

Reika [66]

Answer:

22

Step-by-step explanation:

The standard form for an equation of a circle is $(x-h)^2+(y-k)^2=r^2$ , where h is the x coordinate of the center of the circle, k is the y coordinate, and r is the radius of the circle. Therefore, the radius of this circle squared is 121, meaning that the radius of the circle is 11. Since the diameter of a circle is twice the radius, the radius of this circle is 22. Hope this helps!

8 0

3 years ago

Question 28 (1 point)

aniked [119]

Answer:

+ 16

Step-by-step explanation:

x² - 8x = - 10

to complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 4)x + (- 4) = - 10 + (- 4)²

x² - 8x + 16 = - 10 + 16

thus 16 is added to both sides to complete the square

4 0

2 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