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4 years ago

Simplify 5^2•5^4 A. 5•8 B. 5^8 C. 5^2 D. 5^6

Mathematics

2 answers:

Svetach [21]4 years ago

7 0

D.5^6 you add the 4 and the 2

MakcuM [25]4 years ago

5 0

Add the exponents which leaves you D. 5^6

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9 divided by 234 <br> Breakdown

Hunter-Best [27]

Answer:

0.0384615385

Step-by-step explanation:

look at picture

7 0

4 years ago

I need some help here plz

Svetllana [295]

Answer:

I don't know if this is correct but id say

$5500 \times \frac{2}{100} = 55 \times 2 = 110$

Step-by-step explanation:

the interest is 2 % so you'd say 2/100 times the amount of money in the interest so that is 2% of 5500

the interest is 110

7 0

3 years ago

What is 3/8 divided by 4/5

creativ13 [48]

15/32 hope this helps

6 0

3 years ago

Read 2 more answers

Do the equations x + y = -2 and 3x + 3y = −6 define the same line? Explain.

tino4ka555 [31]

Answer:

Yes, they do.

Step-by-step explanation:

They both pass through the same points of -2 on the x and y axis.

6 0

3 years ago

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0

3 years ago