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Aloiza [94]
3 years ago
15

Rewrite 12*12*12 using exponents

Mathematics
1 answer:
Usimov [2.4K]3 years ago
7 0

Answer:

12^3    PLS GIVE ME BRAINLIEST

Step-by-step explanation:

because u multiply 12 3 times so 12^3

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Solve for the formula for the indicated variable. Solve A=1/2bh for b.
choli [55]
B = 2A/h

The way you would get this is first by clearing the fraction. So in order to clear the fraction, you have to multiply the equation by the denominator.

2(A = 1/2bh)

2A = 1bh

2A = bh

Then you have to isolate the variable. So you'd divide bh by h in order to get b by itself. So you'd end up with:

2A/h = bh/h

2A/h = b
5 0
3 years ago
ABC and ACD are both right angled triangles
emmasim [6.3K]

Answer:

a. Pythagorous theorem

b. 13.9 cm

Step-by-step explanation:

a. By pythagorous theorem,

   AC = \sqrt{12^{2} + 5^{2}} cm = 13 cm

b. By pythagorous theorem,

   AD = \sqrt{13^{2}+5^{2}} cm = 13.9 cm

[If there's any problem with the solution, please inform me]

3 0
4 years ago
Need help with 7 please! I need help this is very hard i tried Everything!!
yulyashka [42]

50/2.5=20 .   20 steps for the average person to travel 50 ft.

5 0
3 years ago
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On the hike down the mountain Erica descends at a rate of 36 speed each hour what is her change in elevation after 3 hours
Korolek [52]

Answer:

C = 108 feet

Step-by-step explanation:

if in 1 hour she descends at a speed of 36

in 3 hours she will descend. x

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6 0
3 years ago
0.6 umol/L to g/L conversion please.
Mamont248 [21]

Answer:

FOR H_2

1 \mu mol/L = 2\times 10^{-6} mol/L

Step-by-step explanation:

Given data:

concentration 0.6 \mu mol/L

WE KNOW THAT

ONE\  \mu = 10^{-6}

therefore

1 \mu mol/L = 10^{-6} mol/L

Now for conversion of mol to gram need the molar mass of element in which particular value convert to.

let us take H_2

we know that molar mass of H_2 is 2 gram

therefore

1 \mu mol/L = 2\times 10^{-6} mol/L

let take other example, for O_2

we know that molar mass of O_2 is 32 gram

therefore

1 \mu mol/L = 32\times 10^{-6} mol/L

5 0
3 years ago
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