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2 years ago

Simplify the expression below. 4^5 x 4^3

Mathematics

2 answers:

BartSMP [9]2 years ago

4 0

Answer:

4^8

Step-by-step explanation:

Because both expressions have the same base, we can add the exponents to simplify the expression:

$4^5*4^3 = 4^{5+3} = 4^8$

elena-s [515]2 years ago

3 0

Answer: 4^8 or 65,536

Step-by-step explanation:

The exponent "product rule" tells us that, when multiplying two powers that have the same base, you can add the exponents.

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Plato combining functions mastery test please help

just olya [345]

Answer:

<em>Choice: B.</em>

Step-by-step explanation:

<u>Operations With Functions</u>

Given the functions:

$f(x)=\sqrt[3]{12x+1}+4$

$g(x)=\log(x-3)+6$

The function (g-f)(x) can be obtained by replacing both functions and subtracting them as follows:

$(g-f)(x)= g(x) - f(x)$

$(g-f)(x)= \log(x-3)+6 - (\sqrt[3]{12x+1}+4)$

Operating:

$(g-f)(x)= \log(x-3)+6 - \sqrt[3]{12x+1}-4$

Joining like terms:

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2 years ago

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Your grandparents invested $2,000 for you on the day you were born. How much will this investment be worth on your 25th birthday

algol [13]

We know that, Amount in Compound interest is given by :

$\bigstar \ \ \boxed{\sf{Amount = Principal\bigg(1 + \dfrac{Rate \ of \ Interest}{100}\bigg)^{Time \ Period}}}$

Given : Principal = $2000

Given : Annual yield is 5% and the interest is compounded quarterly

It means : Interest is compounded 4 times in a year

$\implies \sf{Rate \ of \ Interest = \dfrac{R}{4} = \dfrac{5}{4}}$

$\sf{\implies Time \ period = (25 \times 4) = 100}$

Substituting all the values in the formula, we get :

$\implies \sf{Amount = 2000\bigg(1 + \dfrac{\dfrac{5}{4}}{100}\bigg)^{100}}$

$\implies \sf{Amount = 2000\bigg(1 + \dfrac{5}{400}\bigg)^{100}}$

$\implies \sf{Amount = 2000\bigg(1 + \dfrac{1}{80}\bigg)^{100}}$

$\implies \sf{Amount = 2000\bigg(\dfrac{81}{80}\bigg)^{100}}$

$\implies \sf{Amount = 2000 \times (1.0125)^{100}}$

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$\implies \sf{Amount = 6926.8}$

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2 years ago

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2 years ago

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2 years ago

1. Type an equation in the equation editor that uses 2 fractions with parentheses around one of them. Example: <img src="https:/

avanturin [10]

Answer:

i) $\frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}$ $\Rightarrow$ \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

ii) $4^{3} + 8^{2} + \sqrt{9}$ $\Rightarrow$ 4^{3} + 8^{2} + \sqrt{9}

iii) $(\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}$ (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}

Step-by-step explanation:

i) $\frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}$ $\Rightarrow$ \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

ii) $4^{3} + 8^{2} + \sqrt{9}$ $\Rightarrow$ 4^{3} + 8^{2} + \sqrt{9}

iii) $(\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}$ (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}

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3 years ago