Find the average rate of change between f(-4) and f(-1) in the function f(x)=x^2+2x-8

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

Which pair of inequalities is shown in the graph?

nikitadnepr [17]

Answer:

y > x - 5

and

y > -x + 1

Step-by-step explanation:

Since the solution goes above both the lines, we know that the inequalities must have a y > aspect. To get the next portion, simply find the equation of the lines. This would get you y > x - 5 and y > -x + 1.

4 0

3 years ago

Many non-profit groups ask for donations during December. Do men give higher monetary donations than women in the month of Decem

Lelu [443]

Answer:

For this case we want to test if do men give higher monetary donations than women in the month of December (alternative hypothesis) and the complement should be the null hypothesis. So then the sytem of hypothesis are:

Null hypothesis: $\mu_m -\mu_w \leq 0$

Alternative hypothesis : $\mu_m -\mu_w >0$

So then the best option would be:

d. Ha: mum -muw > 0

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

Solution to the problem

For this case we want to test if do men give higher monetary donations than women in the month of December (alternative hypothesis) and the complement should be the null hypothesis. So then the sytem of hypothesis are:

Null hypothesis: $\mu_m -\mu_w \leq 0$

Alternative hypothesis : $\mu_m -\mu_w >0$

So then the best option would be:

d. Ha: mum -muw > 0

3 0

3 years ago

A) (90 - 24,^3 =<br> Helppppppp

elena55 [62]

Answer:

-13734

Step-by-step explanation:

We're going to use pemdas for this equation.

Since there are no parenthesis we move on to exponents, which we do have, so we would cube 24, which gets us 13,824.

Next we subtract 90 from 13,824 which gives us our answer of -13,734

6 0

3 years ago

Hello , I need help with these 2 problems. Could someone help me ?

saw5 [17]

Step-by-step explanation:

The first one will be:

7x^6 - 7x^5 +28x - 4x² - 60

And the second:

-9x⁴-3x²-4x²+5x

3 0

3 years ago