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Angelina_Jolie [31]
3 years ago
7

Solve |2x-5|=4 pleasee

Mathematics
2 answers:
Anastasy [175]3 years ago
7 0

Answer:

X= 1/2

Step-by-step explanation:

8090 [49]3 years ago
3 0

Answer:

x = 1/2 and x = 9/2

Step-by-step explanation:

To solve this equation: |2x-5|=4 we need two evaluate two cases:

|2x-5| = 2x-5 when x>5/2 ✅

|2x-5| = -2x+5 when x<5/2✅

Then, if x>5/2:

2x-5 = 4 ➡ x = 9/2

Then, if x>5/2:

-2x+5 = 4 ➡ x = 1/2

Then, the two solutions are:  x = 1/2 and x = 9/2

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Answer:

i don't know this type of work.

Step-by-step explanation:

i'm only in 10th grade

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Surface area of cylinders
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Answer:

SA = 402.14 square inches

Step-by-step explanation:

Imagine you unroll the cylinder. You will be left with a rectangle and two "lids" (disks).

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Ar = (12 in) * (8 in) *pi = 301.60 in^2

The area of one disk will be

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3 years ago
Please help me on #12
Verdich [7]
When you rearrange the equation, you find that the solution is b)
3 0
3 years ago
Read 2 more answers
An aquarium is in the shape of a rectangular prism. The area of the base of the aquarium is 250 square inches. Which is the volu
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Step-by-step explanation:

It is given that,

An aquarium is in the shape of a rectangular prism. The area of the base of the aquarium is 250 square inches.

We need to find the volume of the aquarium.

The formula for the volume of the rectangular prism is given by :

V = lbh

Where, l is length, b is width and h is height

Let the height of the prism is h inches. So,

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V = 250× h

= (250h) inch³

By substituting the value of h, we can find its volume.

Hence, the volume of the aquarium is (250h) inch³.

6 0
3 years ago
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Find the radius of convergence and the internval of convergence:
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You can use the root test here. The series will converge if

L=\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{(4-x)^n}{4^n+9^n}}

You have

L=\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{(4-x)^n}{4^n+9^n}}=|4-x|\lim_{n\to\infty}\frac1{\sqrt[n]{4^n+9^n}}

Notice that

\dfrac1{\sqrt[n]{4^n+9^n}}=\dfrac1{\sqrt[n]{9^n}\sqrt[n]{1+\left(\frac49\right)^n}}=\dfrac1{9\sqrt[n]{1+\left(\frac49\right)^n}}

so as n\to\infty, you have \left(\dfrac49\right)^n\to0, which means you end up with

L=\dfrac{|4x-1|}9

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|4x-1|
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