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3 years ago

Solve the inequality. 1/6w≥2.5

2 answers:

7 0

$\dfrac{1}{6}w\geq2.5\qquad\text{multiply both sides by 6}\\\\\not6^1\cdot\dfrac{1}{\not6_1}w\geq6\cdot2.5\\\\\boxed{w\geq15}$

Nostrana [21]3 years ago

5 0

<u><em>Answer: w≥15 *The answer should be positive sign.*</em></u>

Step-by-step explanation:

multiplication property of equality is multiplying both sides of an equation by the same number does not change the equation.

a*c=b*c

multiply by 6 both sides of an equation.

6*1/6w≥2.5*6

simplify by equation.

2.5*6=15

w≥15

Hope this helps!

Thanks!

Have a great day!

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FYI, that L in the denominator is factorial.

This is some pretty serious stuff for high school.

My inclination is that the answer is A, which is the definition of <em>e. </em> Let's see if we can show this.

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3 years ago

Joe and Sue were trying to calculate the square inches of paper needed to create a cone for the cotton candy. The radius

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The correct statements are Neither was correct and Joe used the height and radius to calculate the slant height.

<h3>How to find the square inches of paper needed.</h3>

The square inches of paper needed A = surface area of cone + area of overlap

<h3>Surface area of cone</h3>

The surface area of the cone is given by A = πr[r + √(h² + r²)] where

r = radius of cone = 2 in and
h = height of cone = 6 in.

So, A = πr[r + √(h² + r²)]

A = π × 2 × [2 + √(6² + 2²)]

A = π × 2 × [2 + √(36 + 4)]

A = π × 2 × [2 + √40]

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A = 2π[2 + 2√10]

A = 4π + 4π√10

<h3>The area of overlap</h3>

The area of overlap A' = wL where

w = width of overlap = 1/2 in and
L = slant height of cone = √(h² + r²)

So, A' = wL

A' = w[√(h² + r²)]

A' = 1/2[√(6² + 2²)]

A' = 1/2[√(36 + 4)]

A' = 1/2[√40]

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A' = √10

So, the number of square inches of paper needed is A" = A + A'

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Since the number of square inches of paper needed is 4π + 4π√10 + √10

So, the correct statements are Neither was correct and Joe used the height and radius to calculate the slant height.

Learn more about surface area of cone here:

brainly.com/question/24979679

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