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Mekhanik [1.2K]
4 years ago
9

Solve for e. 4/3=-6e-5/3

Mathematics
1 answer:
cricket20 [7]4 years ago
5 0

Answer:

e = 1/2

Step-by-step explanation:

to solve foe e in  this 4/3=-6e-5/3

solution

4/3=-6e-5/3

4/3 + 5/3 = 6e

find the lcm of the left hand side

4 + 5/3 = 6e

9/3 = 6e

cross multiply

3 x 6e = 9 x 1

18e = 9

divide both sides by the coefficient of e which is 18

18e /18 = 9/18

e = 1/2

therefore the value of e in the expression above is evaluated to be equals to 1/2

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I think A=12bh I hope it’s right
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3 years ago
Fill in the space to complete to equality:<br><br> 3v-v²=v(__)
levacccp [35]

Answer:

3v-v² = v(3-v)

Step-by-step explanation:

Use distributive property on the right side of the equality.

6 0
4 years ago
What's the circumference of a circle with a diameter of 9 inches?​
yaroslaw [1]

Answer:

9pi

Step-by-step explanation:

circumference= 2 (pi) (r)

radius = diameter \ 2

r = 9 \ 2 = 4.5in.

c = 2 (pi) (4.5)

c = 9pi or 28.3in.

Hope this helps!

3 0
3 years ago
Please help!!! Will mark BRAINLIEST!!!!
Bond [772]

In an equilateral triangle, the measure of each angle is 60 degrees. Then the value of x and y is 17/3 and 65.

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

A triangle is a three-sided polygon with three angles. The angles of the triangle add up to 180 degrees.

In trianlge ΔMNP, PM = 3x - 4, MN = 14, ∠N = 5y - 5.

In an equilateral triangle, all the sides and all the angles are equal.

MN = NP = PM

∠M = ∠N = ∠P

Then we have

   PM = MN

3x - 4 = 14

      x = 17/3

We know in an equilateral triangle, the measure of each angle is 60 degrees. Then we have

    ∠N = 60

5y - 5 = 60

    5y = 65

      y = 13

More about the triangle link is given below.

brainly.com/question/25813512

#SPJ1

6 0
2 years ago
Need help on this question too hard :((​
andrew-mc [135]

Answer:

2/3

Step-by-step explanation:

Convert the fractions to decimals.

2/3 = 0.666

4/5 = 0.8

14/18 = 0.777

34/50 = 0.68

From the decimals, 0.666 is the only option which falls between 0.5 and 0.75.

5 0
3 years ago
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