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levacccp [35]
3 years ago
8

Find the equation. Plz.

Mathematics
2 answers:
viktelen [127]3 years ago
5 0

Answer:

Answer is B, y=x+67

Step-by-step explanation:

Each x-value (left numbers) will become the number on the right when added to 67. The numbers next to the x's in the answer choices represents multiplication to get the y-values, and since there is no multiplication, just the addition of 67, the answer is y= x+67

Hope this helps!

Kobotan [32]3 years ago
3 0

the answer is

A) y=5x+67

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A palindrome is a number that reads the same forwards and backwards, such as 22 or 39793.
Aleks04 [339]

Answer:

There are thirty-nine palindromic numbers between 10 and 400.

Step-by-step explanation:

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5 0
2 years ago
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
3 years ago
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