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ArbitrLikvidat [17]
3 years ago
13

Simplify the following expression:

rt{x} 48a^{8}" alt="\sqrt{x} 48a^{8}" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
fenix001 [56]3 years ago
4 0

Answer:

48\sqrt{x}a^{8}

Step-by-step explanation:

\sqrt{x}48a^{8}

rewrite using communitive property of multiplication:

48\sqrt{x}a^{8}

Andreyy893 years ago
3 0

\framebox{\parbox[t][1.0cm]{4.50cm}{\addvspace{0.2cm} \centering $  \;\sqrt[48]{xa^8} \; \, \dfra\;\   \;\; \dfrac{}{} $ } }\\

Step-by-step explanation:

\sqrt{x48a^8}

There are no variables we will just rewrite this as:

\sqrt[48]{xa^8}

Hope this helps!

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4w+15= -13<br><br> -12=n/2-6
Firlakuza [10]

Answer:

For the one including 4w the answer is - 7

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A soccer ball is kicked from 3 feet above the ground. the height,h, in feet of the ball after t seconds is given by the function
dexar [7]
Here's our equation.

h=-16t+64t+3

We want to find out when it returns to ground level (h = 0)

To find this out, we can plug in 0 and solve for t.

0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a}  \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16}  = \frac{64\±\sqrt{4096+192}}{32}

= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}

So the ball will return to the ground at the positive value of \boxed{\frac{8+\sqrt{67}}{4}} seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2

h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}


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Ivanshal [37]
To add, subtract, multiply or divide the number given to get a whole number or an integer
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