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1 year ago

X2 − 2x − 2 into the form a(x − h)2 + k.

Mathematics

1 answer:

Sedbober [7]1 year ago

3 0

Answer:

x² - 2x - 2

By Completing the Square....

x² - 2x + 1² - 2 - 1²

(x - 1)² - 3 .....

Hope it helps.

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Based on the line plot, how many recipes call for more than 1/4 tsp of salt?

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It would be 6 haha I just took a test and got it right

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2 years ago

Read 2 more answers

Divide these polynomials. (8x³ + 2x² - 7x + 2) ÷ (2x - 1)

Novay_Z [31]

Answer:

The answer is letter A.

(8x³ + 2x² - 7x + 2) ÷ (2x - 1)

8x³ + 2x² - 7x + 2 ÷ 2x - 1

(2x-1)(4x²+3x-2)÷2x-1

cancel 2x-1 so that the answer is 4x²+3x-2.

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2 years ago

Write an equation of the line below

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Answer:

y=1/2+1

Step-by-step explanation:

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2 years ago

A community group is picking up trash along a portion of the highway that is 1.48 miles long. In one hour they can clean a secti

aleksklad [387]

Answer:

12 if im correct

Step-by-step explanation:

first we have to see that they are doing 1/8 of a mile per hour so if we divide 1.48 by 1/8 it should give us the time

first we convert 1/8 to a decimal which is .125 and we divide 1.48 by .125

giving us 11.84

since it is time, 11 hrs would be too little of a time so we need to round up to 12

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2 years ago

Read 2 more answers

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

2 years ago