How do you simplify radical and rational exponents?

coldgirl [10]

Answer:

x=3

Step-by-step explanation:

Use pemdas!

First use distributive property to get rid of the parenthesis, by multiplying 2 by 3x and -4. Then combine like terms and use inverse operations to simplify each side until you get it down to just x! Hope this helped! :)

3 0

3 years ago

REPOST 100 POINTS+BRAINLIEST 8th grade math- counting principle - 7 short questions

madreJ [45]

I only have one for 12
12, 21
24, 42
36, 63
48, 84

7 0

1 year ago

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A new car is purchased for 24600 dollars. The value of the car depreciates at 5.5% per year. To the nearest year, how long will

boyakko [2]

Answer:

The number of years until the value of the car is 11300 dollars is 9 years.

Step-by-step explanation:

This can be calculated using the following formula:

Value of the Car = Purchase cost - (Annual depreciation expenses * N) ........ (1)

Where;

N = Number of years until the value of the car is 11300 dollars

Purchase cost = $24,600

Annual depreciation rate = 5.5%

Annual depreciation expenses = $24,600 * 5.5% = $1,353

Value of the car = $11,300

Substituting the value into equation (1) and solve for N, we have:

$11,300 = $24,000 - ($1,353 * N)

$1,353 * N = $24,000 - $11,300

$1,353 * N = $12,700

N = $12,700 / $1,353

N = 9.38654841093865

Rounding to the nearest year as required by the question, we have:

N = 9

Therefore, the number of years until the value of the car is 11300 dollars is 9 years.

Step-by-step explanation:

4 0

3 years ago

Express in terms of logarithm without exponents

sdas [7]

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)
\\ \quad \\
% Logarithm of rationals
\\ \quad \\
% Logarithm of exponentials
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-----------------------------\\\\
log_b(xy^2z^{-6}\implies log_b(x)+log_b(y^2)+log_b(z^{-6})
\\\\\\log_b(x)+2log_b(y)-6log_b(z)$

now, the one below that, which is equivalent to that? well, just look above it

6 0

3 years ago

the pitch, or frequency, of a vibrating string varies directly with the square root of the tension. if a string vibrates at a fr

Naily [24]

$\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\
\begin{array}{llll}
\stackrel{frequency}{f}\textit{ of a vibrating string varies directly}\\
\qquad \qquad \textit{with the square root of the tension }\stackrel{tension}{t}
\end{array}$

$\bf f=k\sqrt{t}\quad \textit{we also know that }
\begin{cases}
f=300\\
t=8
\end{cases}\implies 300=k\sqrt{8}
\\\\\\
\cfrac{300}{\sqrt{8}}=k\implies \cfrac{300}{\sqrt{2^2\cdot 2}}=k\implies \cfrac{300}{2\sqrt{2}}=k\implies \cfrac{150}{\sqrt{2}}=k
\\\\\\
\textit{and we can \underline{rationalize} it to }\cfrac{150\sqrt{2}}{2}\implies 75\sqrt{2}=k$

$\bf thus\qquad f=\stackrel{k}{75\sqrt{2}}\sqrt{t}\implies \boxed{f=75\sqrt{2t}}\\\\
-------------------------------\\\\
\textit{now, when t = 72, what is \underline{f}?}\qquad f=75\sqrt{2(72)}$

5 0

3 years ago

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