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2 years ago

5

Geometry help please I'm stuck will give brainiest answer and thanks and five stars if correct

1 answer:

Tpy6a [65]2 years ago

8 0

1. Reduction: 1/2
2. 200 millimeters
3. Glide Reflection

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Which is the value of the expression (StartFraction (10 Superscript 4 Baseline) (5 squared) Over (10 cubed) (5 cubed)) cubed?

Flura [38]

Answer:

The value to the given expression is 8

Therefore $\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=8$

Step-by-step explanation:

Given expression is (StartFraction (10 Superscript 4 Baseline) (5 squared) Over (10 cubed) (5 cubed)) cubed

Given expression can be written as below

$\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3$

To find the value of the given expression:

$\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=\frac{((10^4)(5^2))^3}{((10^3)(5^3))^3}$

( By using the property ( $(\frac{a}{b})^m=\frac{a^m}{b^m}$ )

$=\frac{(10^4)^3(5^2)^3}{(10^3)^3(5^3)^3}$

( By using the property $(ab)^m=a^mb^m$ )

$=\frac{(10^{12})(5^6)}{(10^9)(5^9)}$

( By using the property $(a^m)^n=a^{mn}$ )

$=(10^{12})(5^6)(10^{-9})(5^{-9})$

( By using the property $\frac{1}{a^m}=a^{-m}$ )

$=(10^{12-9})(5^{6-9})$ (By using the property $a^m.b^n=a^{m+n}$ )

$=(10^3)(5^{-3})$

$=\frac{10^3}{5^3}$ ( By using the property $a^{-m}=\frac{1}{a^m}$ )

$=\frac{1000}{125}$

$=8$

Therefore $\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=8$

Therefore the value to the given expression is 8

3 0

2 years ago

Read 2 more answers

34/23 as a decimal rounded to the nearest tenth

Ainat [17]

Answer:

1.5

Step-by-step explanation:

1.47826086

rounded to the nearest tenth is 1.5

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2 years ago

What is the sum of /-10/ and /6/

kap26 [50]

Answer:

16

Step-by-step explanation:

-10 opposite is 10 which this \ or | means the its positive number. So 10 +6 is 16

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2 years ago

A polynomial function g(x) has a zero of 8-5i with multiplicity 4. What is the minimum possible degree of f(x)?

Nikitich [7]

I think it’s B hope it’s right

8 0

2 years ago

Please help +25 points

mestny [16]

The answer is D.
Hope I helped

8 0

1 year ago