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mote1985 [20]
2 years ago
15

What expressions are equivalent to the one below. check all that apply. 8^x

Mathematics
1 answer:
tangare [24]2 years ago
7 0

The (32/4)^x is equal to the 8^x.

We have given that,

A. (32/4)^x

B. 32^x/4

C. 8*8^x+1

D. x^4

E. 32^x/4^x

F. 8*8^x-1

We have to determine the expression equivalent to the 8^x.

<h3>What is the expression?</h3>

An expression or mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context.

Therefore the expression equivalent to the 8^x is A.

That is,

=\frac{32}{4}^x\\ =8^x

Therefore the (32/4)^x is equal to the 8^x.

To learn more about the expression visit:

brainly.com/question/723406

#SPJ1

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In Hazzard County, 57 of the schools have a teacher-to-student ratio that meets or exceeds the requirements for accreditation. T
Tcecarenko [31]

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3:2

Step-by-step explanation:

8 0
3 years ago
|20x|=2|2x+8|<br> please show all steps <br> and both answers
lukranit [14]

Step-by-step explanation:

20x = 4x +16

16x = 16

x=1

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7 0
4 years ago
Jorge is asked to build a box in the shape of a rectangular prism. The maximum girth of the box is 20 cm. What is the width of t
MariettaO [177]

Answer:

The width of the box is 6.7 cm

The maximum volume is 148.1 cm³

Step-by-step explanation:

The given parameters of the box Jorge is asked to build are;

The maximum girth of the box = 20 cm

The nature of the sides of the box = 2 square sides and 4 rectangular sides

The side length of square side of the box = w

The length of the rectangular side of the box = l

Therefore, we have;

The girth = 2·w + 2·l = 20 cm

∴ w + l = 20/2 = 10

w + l = 10

l = 10 - w

The volume of the box, V = Area of square side × Length of rectangular side

∴ V = w × w × l = w × w × (10 - w)

V = 10·w² - w³

At the maximum volume, we have;

dV/dw = d(10·w² - w³)/dw = 0

∴ d(10·w² - w³)/dw = 2×10·w - 3·w² = 0

2×10·w - 3·w² = 20·w - 3·w² = 0

20·w - 3·w² = 0 at the maximum volume

w·(20 - 3·w) = 0

∴ w = 0 or w = 20/3 = 6.\overline 6

Given that 6.\overline 6 > 0, we have;

At the maximum volume, the width of the block, w = 6.\overline 6 cm ≈ 6.7 cm

The maximum volume, V_{max}, is therefore given when w = 6.\overline 6 cm = 20/3 cm  as follows;

V = 10·w² - w³

V_{max} = 10·(20/3)² - (20/3)³ = 4000/27 = 148.\overline {148}

The maximum volume, V_{max} = 148.\overline {148} cm³ ≈ 148.1 cm³

Using a graphing calculator, also, we have by finding the extremum of the function V = 10·w² - w³, the coordinate of the maximum point is (20/3, 4000/27)

The width of the box is;

6.7 cm

The maximum volume is;

148.1 cm³

5 0
3 years ago
What additional information could be used to prove ΔABC ≅ ΔMQR using SAS? Select two options. m∠A = 64° and AB = MQ = 31 cm CB =
enyata [817]

Answer:

m∠R = 60° and AB ≅ MQ

m∠Q = 56° and CB ≅ RQ

Step-by-step explanation:

Given data :

Prove ΔABC ≅ ΔMQR using SAS

The  missing information to prove ΔABC ≅ ΔMQR using SAS

  • m∠R = 60° and AB ≅ MQ
  • m∠Q = 56° and CB ≅ RQ
3 0
3 years ago
Read 2 more answers
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