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Mathematics

1 answer:

atroni [7]3 years ago

5 0

The answer is B there you go have a nice day

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Which expression is equivalent to ?

mamaluj [8]

Answer:

Step-by-step explanation:

$\frac{(a^{2}b^{4}c)^{2}*(6a^{3}b)*(2c^{5})^{3}}{4a^{6}b^{12}c{^3}}\\\\=\frac{a^{2*2}b^{4*2}c^{2}6a^{3}b2^{3}c^{5*3}}{4a^{6}b^{12}c{^3}}\\\\=\frac{a^{4}b^{8}c^{2}6a^{3}b8c^{15}}{4a^{6}b^{12}c{^3}}\\\\=\frac{6*8*a^{4+3}b^{8+1}c^{2+15}}{4a^{6}b^{12}c{^3}}\\\\=\frac{6*2*a^{7}b^{9}c^{17}}{a^{6}b^{12}c{^3}}\\\\=\frac{12a^{7-6}c{17-3}}{b^{12-9}}\\\\=\frac{12a^{1}c^{14}}{b^{3}}\\\\=\frac{12ac^{14}}{b^{3}}$

Hint:

$x^{m}*x^{n}=x^{m+n}\\\frac{x^{m}}{x^{n}}=x^{m-n},m>n\\\frac{x^{m}}{x^{n}}=\frac{1}{x^{n-m}},n>m$

5 0

3 years ago

I need help urgent plz someone help me solved this problem! Can someone plz help I’m giving you 10 points! I need help plz help

VMariaS [17]

Answer: a) 8.779 years

b) 8.664 years

<u>Step-by-step explanation:</u>

$A=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}$

A: accumulated amount (balance)
P: principal amount (original/initial investment)
r: interest rate (convert to a decimal)
n: number of times compounded per year
t: number of years

Given: A = 1800, P = 900, r = 8% = 0.08, n = 3, t = unknown

$1800=900\bigg(1+\dfrac{0.08}{3}\bigg)^{3t}\\\\\\2=\bigg(1+\dfrac{0.08}{3}\bigg)^{3t}\\\\\\ln\ 2=ln \bigg(1+\dfrac{0.08}{3}\bigg)^{3t}\\\\\\ln\ 2=3t\ ln\bigg(1+\dfrac{0.08}{3}\bigg)\\\\\\\dfrac{ln\ 2}{3\ ln\bigg(1+\dfrac{0.08}{3}\bigg)}=t\\\\\\\large\boxed{8.779=t}$