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2 years ago

the ratio of cars to trucks at an auto dealer is 3/2. if there are 144 cars at the dealership, how many trucks are there?

Mathematics

1 answer:

lana [24]2 years ago

5 0

Answer:

Step-by-step explanation:

for every 3 cars there is 2 trucks

If you have 144 cars then we divide 144 by 3 to get 48

That is not the answer though because that would be a 3/1 ratio

We then multiply 48 by 2 to get 96

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1 year ago

A vat with 700 gallons of beer contains 4% alcohol (by volume). Beer with 6% alcohol is pumped into the vat at a rate of 7 gal/m

dolphi86 [110]

Answer:

5.2%.

Step-by-step explanation:

After 1 hour (60 / 100) * 700 = 420 gallons of beer have been exchanged.

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So the strength at that time = (420* 6) + 280*4) / 700

= 5.2%.

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2 years ago

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2 years ago

A sequence is defined by the recursive function f(n+1)=f(n). If f(3)=9, what is f(1)?

vekshin1

Put n=2 ,
f(2+1) = f(2) => f(3) = f(2) = 9
now, put n=1,
f(1+1) = f(1) => f(2) = f(1) = 9

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2 years ago

Read 2 more answers

"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

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

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

7 0

2 years ago

the ratio of cars to trucks at an auto dealer is 3/2. if there are 144 cars at the dealership, how many trucks are there?​

the ratio of cars to trucks at an auto dealer is 3/2. if there are 144 cars at the dealership, how many trucks are there?