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BigorU [14]
3 years ago
10

Is this question correct?

Mathematics
1 answer:
Andrews [41]3 years ago
6 0

Answer:

10^20

3^11

3^24

10^12

Step-by-step explanation:

(10^2)^10=10^2*10=10^20

3^3*3^8=3^3+8=3^11

(3^3)^8=3^3*8=3^24

10^2*10^10=10^2+10=10^12

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If your parallelogram has a base a 60 degrees what is your height
DENIUS [597]
I really don’t know I’m just tryna get answers bc I’m failing and stuff and I’m sorry but ima just eh
3 0
3 years ago
Can someone PLEASE explain how to simplify square roots with variables and eexponents in them?? I'd also be thankful if you expl
Verizon [17]
X^a/b is  \sqrt[b]{x^a} . The way I memorise that is x^1/3 is the cubic root of x. Do you get it? In that case, x is raised to a power of 1 and the cubic root is practically has a power of 3.
In your example, 

\sqrt[ \frac{3}{2} ]{16 x^4} is practically square rooting each term then cubing them individually. Remember when square-rooting any index you halve it. I'll elaborate:

\sqrt{x^4} = x^{2}
\sqrt{16} = 4
Then cube each,
4^3 = 64 
and ( x^{2} )^3 = x^{6}

As for the 2nd part: you must use the rules of indices.
x^{a}  *  x^{b} =  x^{a+b}
So breaking the question up:

3 * 3 = 9
x^{ \frac{1}{2} } stays as is since the 2nd term does not contain x
now: 
y^{ \frac{4}{3} }  * y^{1} =  y^{ \frac{4}{3} + 1 }  =  y^{ \frac{4}{3} +  \frac{3}{3} }  =   y^{ \frac{7}{3} }
This makes your final answer look like this:
9 x^{ \frac{1}{2} }  y^{ \frac{7}{3} }

I hope that helped and good luck in your test!
4 0
4 years ago
Jane gets 17/20 apples. Lemm wants the same amount but a different way. How much should she get and what strategy should she use
Elena-2011 [213]

Answer:

34/40

????????

Step-by-step explanation:

you multiply them both by two??

6 0
3 years ago
F(5)=-3x+10 what will the domain and range be
Vladimir [108]
F= -3/5 x+2. Hope this helps

3 0
3 years ago
Can I get the answers for number 14 plz?
nataly862011 [7]
A.
-1: 
2(2)^{-1}  \\ 2( \frac{1}{2}) \\  \frac{2}{2}  \\ 1
(-1,1)

0:
2(2)^{0} \\ 2(1) \\ 2
(0, 2)

1:
2(2)^{1}  \\ 2(2) \\ 4
(1, 4)

2:
2(2)^{2} \\ 2(4) \\ 8
(2, 8)

3:
2(2) ^{3}  \\ 2(8) \\ 16
(3,16)

b.
To graph the equation, simply go through the points (-2, 0.5), (-1, 1), (0,2), (1,4), (2,8), and (3,16). Make sure you never go below 0 on the x-axis, because there's an asymptote there.

Hope this helps!
8 0
3 years ago
Read 2 more answers
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