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2 years ago

Solve and make a table ( you may decide what the topic is and the value is.Algebra I)

Mathematics

2 answers:

ratelena [41]2 years ago

6 0

Answer:

Hello...

I'm new user here.

Can you teach me how to use brainly? Please, it would bey favour.

I have just downloaded the app and don't know how to use.

lora16 [44]2 years ago

3 0

I really don’t know I just need points

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(4000/2500)100=160%

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3 years ago

Select the correct answer.

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F(x)=2^x is the answer

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3 years ago

A tablet weighs 1.23 pounds. What is its weight written as a mixed number? Plus workings

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3 years ago

Two consecutive integers have a sum of 153. find the integers.

Natasha2012 [34]

X = first integer
x + 1 = second consecutive integer

x + (x + 1) = 153..combine like terms
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3 years ago

I’m lost again can somebody help asap ??

serious [3.7K]

Fully detailed and equipped information process right below. The answer for this question is "25" (simplified form).

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(5^{- 5} \times 2^8 \times 1 \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(2^8 \times 1 \times \dfrac{1}{5^5} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{2^8 \times 1}{5^5} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{256}{3125} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(\dfrac{3125}{256} \Bigg)^2}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1}{2^3 \times 5^2} \Bigg)^4 \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(2^3 \times 5^2)^4} \Bigg) \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(5^2)^4 \times (2^3)^4} \Bigg) \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \dfrac{1}{2^{12} \times 5^8} \times \dfrac{3125^2}{256^2}}$

$\mathbf{\dfrac{3125^2 \times 1 \times 2^{28}}{256^2 \times 5^8 \times 2^{12}}}$

$\mathbf{\dfrac{2^{(28 - 12)} \times 3125^2}{256^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 3125^2}{256^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times (5^5)^2}{(2^8)^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 5^{10}}{2^{16} \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 5^{10 - 8}}{2^{16}}}$

$\mathbf{\dfrac{5^{10 - 8}}{1}}$

$\mathbf{\therefore \quad 5^2}$

$\mathbf{\therefore \quad 25}$

$\boxed{\mathbf{\underline{\therefore \quad Final \: \: Answer \: \: is: \: 25}}}$

Hope it helps.

7 0

3 years ago