All categories

2 years ago

Sally is three times as old as Sam. The sum of their ages is 84. How old is Sally?

Mathematics

1 answer:

Pepsi [2]2 years ago

4 0

Answer:

Sally is 63

Step-by-step explanation:

Represent Sally's age by x and Sam's by y.

Then x = 3y and x + y = 84.

Substituting 3y for x in the second equation, we get:

3y + y = 84, or 4y = 84, or y = 21. Since Sally is 3 times as old as Sam, we multiply 21 by 3: Sally's age is 3(21 years) = 63 years

You might be interested in

In desperate need of help, please help. PLEASE. If possible send work, so I could figure out how to do it.! -Jo:)

podryga [215]

If you have any questions let me know. :)

8 0

2 years ago

What is the equation in slope intercept form?

Klio2033 [76]

Answer:

y=-x+3

Step-by-step explanation:

rise/run

-1/1

-1

4 0

3 years ago

Subtract -5x^2+7x+1 from 8x^2+x-10

melomori [17]

13x^2 + 8x - 9
I hope this helps

4 0

3 years ago

Read 2 more answers

Can someone help me out if question 37. triangle is a SSS, SAS or ASA? ASAP, gotta have an answer fast pls

VARVARA [1.3K]

Answer:

ASA Postulate

Step-by-step explanation:

The two angles and the side included between them of one triangle are equal to the corresponding angles and the included side of the other triangle. (Angle, Common Side, Angle)

3 0

1 year 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

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

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

7 0

2 years ago