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3 years ago

Choose the simplified form of the fifth term of the expansion. -60x2y8 -30x2y8 30x2y8 60x2y8

Mathematics

2 answers:

alexdok [17]3 years ago

8 0

Answer:

Step-by-step explanation:

i just did it

Vika [28.1K]3 years ago

5 0

Answer:

Answer is D on edge

Step-by-step explanation:

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Im distributive property and combining like term and I have to simplify this expression 10(7r)

faltersainse [42]

It would be 70r if you are using distributive property

8 0

3 years ago

Banabas must pay his ex-wife an amount of R350 000 in two years’ time. Calculate the amount that he must invest today to have th

White raven [17]

Answer:

He must invest R297 521 today.

Step-by-step explanation:

The compound interest formula is given by:

$A(t) = P(1 + \frac{r}{n})^{nt}$

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.

Banabas must pay his ex-wife an amount of R350 000 in two years’ time.

This means that $t = 2, A(t) = 350000$

Interest rate of 8.15% per annum compounded monthly:

This means that $r = 0.0815, n = 12$ .

Amount he must invest today:

This is P. So

$A(t) = P(1 + \frac{r}{n})^{nt}$

$350000= P(1 + \frac{0.0815}{12})^{2*12}$

$P = \frac{350000}{(1 + \frac{0.0815}{12})^{2*12}}$

$P = 297521$

He must invest R297 521 today.

4 0

3 years ago

Can someone help me ?

prisoha [69]

I’m pretty sure there are infinite solutions as the system of equations are equivalent to each other.

7 0

3 years ago

Read 2 more answers

Mr. And Mrs ikan deposited $15,000 into college savings accounts with an annual compound interest rate 5.25% when their daughter

m_a_m_a [10]

Answer:

$37,650

Step-by-step explanation:

A=15000(1+0.0525)^18

A=15000(2.51)

A=37650

8 0

3 years ago

Limit

STatiana [176]

Rationalize the numerator:

$\dfrac{\sqrt{x+4}-2}x\cdot\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\dfrac{(\sqrt{x+4})^2-2^2}{x(\sqrt{x+4}+2)}=\dfrac x{x(\sqrt{x+4}+2)}=\dfrac1{\sqrt{x+4}+2}$

This is continuous at $x=0$ , so we can evaluate the limit directly by substitution:

$\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-2}x=\lim_{x\to0}\frac1{\sqrt{x+4}+2}=\frac1{\sqrt4+2}=\frac14$

5 0

3 years ago