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3 years ago

15

During a promotional weekend, a state fair gives free admission to every 155th person that enters the fair. On saturday, there w

ere 5,516 people attending the fair. On sunday, there were 6,399 people there. How many people received a free admission over the two days?

1 answer:

lara31 [8.8K]3 years ago

4 0

76 free admisions where given out

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The ball's height (in meters above the ground), xxx seconds after Antoine threw it, is modeled by: h(x)=-2x^2+4x+16 How many sec

zlopas [31]

Answer:

Step-by-step explanation:

If we are given the quadratic model, we know that the height of anything on the ground is 0 meters, so that model becomes

$0=-2t^2+4t+16$ I changed your x's to t's since t is for time.

Now we need to factor that quadratic and solve for time:

$0=-2(t^2-2t-8)$

Since -2 definitely does not equal 0, then

$t^2-2t-8=0$

Factor that and find t values of 4 and -2. Again, we know that time will never be negative, so

t = 4 seconds.

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3 years ago

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What is the answer to 3s=45?

german

S= 15

3s=45
3s/3=45/3
S=15
You have to divide both sides

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3 years ago

Equivalent ration of <br> b/40=9/10<br> b=____ ?

NARA [144]

Answer:

The correct answer is <em>b = 36</em>

Step-by-step explanation:

In order to find out what b is, you must find out what 9/10 is.

9/10 is 0.9. This will later be used to check our answer.

Now, solve by using cross-multiplication

b * 10 = 9 * 40.

Then use the commutative property to reorder the terms

10b = 9 * 40

Now Multiply 9 * 40 = 360

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Now divide both sides of the equation by 10

<em>b = 36</em>

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3 years ago

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

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3 years ago

Positive negative or no correlation?

bixtya [17]

Answer:

positive correlation

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3 years ago

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