Please answer my question!! Will give brainliest, I beg of you!!

Mathematics

1 answer:

fgiga [73]3 years ago

5 0

Your answer should be 93 because 7 times 7.5 is 52.5 and 9 times 4.5 is 40.5 40.5+52.5=93!

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Find the square: (5x - 4y)2 <br><br> (and please put the process of solving the equation)

satela [25.4K]

$(5x-4y)^2$
$(5x-4y)(5x-4y)$

Multiply each term in one bracket in turn by the terms in the other bracket

$(5x)(5x) = 25x^2$
$(5x)(-4y)=-20xy$
$(-4y)(5x)=-20xy$
$(-4y)(-4y)=16y^2$

Put them together we have
$25x^2-20xy-20xy-16y^2$
$25x^2-40xy-16y^2$

7 0

3 years ago

On a field trip they take busses and vans. Each bus holds 60 people and each van holds 10 people. If 390 people went on the trip

riadik2000 [5.3K]

Answer:

6 buses and 3 vans

Step-by-step explanation:

5 0

3 years ago

Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1

defon

I'm guessing the series is supposed to be

$\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}$

By the ratio test, the series converges if the following limit is less than 1.

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|$

The first $n$ terms in the numerator's denominator cancel with the denominator's denominator:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|$

$|x^n|$ also cancels out and the remaining factor of $|x|$ can be pulled out of the limit (as it doesn't depend on $n$ ).

$\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0$

which means the series converges everywhere (independently of $x$ ), and so the radius of convergence is infinite.