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Olenka [21]
2 years ago
13

A painter is painting a rectangular wall that measures 20.5 ft by 16.4 ft. Each gallon can of paint covers 200 square feet. What

are the fewest gallons needed to paint the wall?
Mathematics
2 answers:
alekssr [168]2 years ago
5 0

The painter needs at least two gallons of paint to cover the wall.

To get this answer, you first need to find the area of the wall; by multiplying 20.5 x 16.4. This will get you 336.2.

Then, you need to divide 336.2 by 200. This will get you approx. 1.6.

If it's asking for partial gallons as well, stop here. Your answer is 1.6.

If not, continue.

Unless you're including partial gallons, you can't get 1.6 gallons of paint, so you need to round up. This will be 2 gallons.

PSYCHO15rus [73]2 years ago
3 0

Answer:

2

Step-by-step explanation:

First calculate the area of the wall.

Area of a rectangle = width x length

Width = 16.4 ft and length = 20.5 ft

Therefore, area = 16.4 x 20.5 = 336.2 ft²

If one gallon can of paint covers 200 ft²  then we need 2 cans of paint to paint a wall measuring 336.2 ft² (as 2 cans of paint = 2 x 200 = 400 ft²)                          

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Answer:

x≥4

Step-by-step explanation:

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group like terms

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3 years ago
Endpoint 1: (-1,2)<br> Midpoint: (9,-6)<br> Endpoint 2=
Vlad [161]

Answer:

(19 , -14)

Step-by-step explanation:

Find the distance in between each x & y for a coordinate.

Let: (x₁ , y₁) = (-1 , 2)

Let: (x₂ , y₂) = (9 , -6)

From x₁ ⇒ x₂: 9 - (-1) = 10

From y₁ ⇒ y₂: -6 - 2 = -8 = 8*

*Remember that distance cannot be negative, but for the sake of this question, we will leave it as -8.

The distance between the x points are in intervals of 10. The distance between the y points are in intervals of 8. Add 10 & subtract 8 to their respective numbers to get endpoint 2:

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8090 [49]

We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!

We are given with ;

{:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}

Take LCM on both sides :

{:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}

<em>Multiplying</em> both sides by <em>10</em> ;

{:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}

{:\implies \quad \sf 165+6y=7y+150}

Can be <em>further written</em> as ;

{:\implies \quad \sf 7y+150=165+6y}

Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>

{:\implies \quad \sf 7y-6y=165-150}

{:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}

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