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Firdavs [7]
3 years ago
12

Ashwin helps paint a square mural in his classroom. Then he helps paint a mural in the hallway whose length is 7 feet longer and

whose width is 3 feet shorter. Let x represent the side length of the mural in ashwins classroom. Write an expression that represents the two binomials you would multiply to find the area of the hallway mural. Use that expression to find the area of the hallway mural. Use that expression to find the area of the hallway mural if each side of the classroom mural is 8 foot long.
Mathematics
2 answers:
melamori03 [73]3 years ago
4 0

Answer: A(x+7)(x-3); 75 square feet

Step-by-step explanation:

Right Answer for apx math

Serhud [2]3 years ago
3 0

Ashwin paints 2 murals:

  1. A square mural in his classroom
  2. A mural in the hallway with length 7 feet longer and width 3 feet shorter

Also, x indicates the side length of the mural in the classroom. Since the classroom mural is square in shape, the length and breadth of the mural would be x.

Part A: Expression to represents the two binomials to be multiplied to find the area of the hallway mural

Area of the hallway mural = Length of the hallway mural × Breadth of the hallway mural

So Length of the hallway mural = Length of the classroom mural + 7

⇒ Length of the hallway mural = x + 7

Breadth of the hallway mural = Breadth of the classroom mural - 3

⇒ Breadth of the hallway mural = x - 3

Hence, x + 7 and x - 3 need to be multiplied to determine the area of the mural

Part B: Area of the hallway mural

Area of the hallway mural = Length of the hallway mural × Breadth of the hallway mural

⇒ Area of the hallway mural = (x + 7) × (x-3)

⇒ Area of the hallway mural = x² + 7x - 3x - 21

⇒ Area of the hallway mural = x² + 4x - 21

Part C: Area of the hallway mural if each side of the classroom mural is 8 foot long

Putting the value x = 8 in the answer of Part B

⇒ Area of the hallway mural = 8² + 4×8 - 21

⇒ Area of the hallway mural = 64 + 4×8 - 21

⇒ Area of the hallway mural = 64 + 32 - 21

⇒ Area of the hallway mural = 75 square feet


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Math be hard- Can ya help?
horrorfan [7]

Answer:

1/8

Step-by-step explanation:

The slope equation for this problem is

y2-y1

--------

x2-x1

So the y2=4, y1=3, x2=8, and x1=0

4-3      =1

-----       --

8-0     =8

Which is 1/8

7 0
2 years ago
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Why is 5/3 greater than 5/8
vazorg [7]
Since 5/3 is an inproper fraction it then becomes 1 2/3 and if you put both of the fractions into a common denominator you get 1 16/24>15/24
Please give me a thanks if you feels it is the right answer
7 0
3 years ago
(I've been trying to figure this out for 3 days and I really need help)
liq [111]

Check the picture below.

since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.

1)

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7

2)

now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.

\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55

3)

well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.

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pretty much the same thing, we get the volume of the cone and its top, add them up.

\bf \stackrel{\textit{cone's volume}}{\cfrac{\pi (3)^2(8)}{3}}~~~~+~~~~\stackrel{\stackrel{\textit{half a sphere}}{\textit{top's volume}}}{\cfrac{4\pi 3^3}{3}\div 2}\implies 24\pi +18\pi \implies 42\pi ~~\approx~~131.95~in^

8 0
3 years ago
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sergeinik [125]
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8 0
3 years ago
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An expression is shown below: 2x3y + 18xy − 10x2y − 90y Part A: Rewrite the expression by factoring out the greatest common fact
SCORPION-xisa [38]

Answer:

Part A : 2y( x³ + 9x - 5x² - 45 ), Part B : 2y( x - 5 )( x² + 9 )

Step-by-step explanation:

Part A : Let's break every term down here to their " prime factors ", and see what is common among them,

2x³y + 18xy − 10x²y − 90y -

2x³y  = 2 * x³ * y,

18xy = 2 * 3 * 3 * x * y,

− 10x²y = 2 * - 5 * x² * y, - so as you can see for this example I purposely broke down - 10 into 2 and - 5. I could have placed the negative on the 2, but as that value was must likely common among all the terms, I decided to place it on the 5. The same goes for " − 90y. " I placed the negative there on the 5 once more.

− 90y = 2 * - 5 * 3 * 3 * y

The terms common among each term are 2 and y. Therefore, the GCF ( greatest common factor ) is 2x. Let's now factor the expression using this value.

2y( x³ + 9x - 5x² - 45 )

Part B : Let's simply factor this entire expression. Of course starting with the " factored " expression : 2y( x³ + 9x - 5x² - 45 ),

2y\left(x^3+9x-5x^2-45\right) - Factor out " (x^3+9x-5x^2-45\right)) " by grouping,

\left(x^3-5x^2\right)+\left(9x-45\right) - Factor 9 from 9x - 45 and x² from x³ - 5x²,

9\left(x-5\right)+x^2\left(x-5\right) - Factor out common term x - 5,

\left(x-5\right)\left(x^2+9\right) - And our solution is thus 2y( x - 5 )( x² + 9 )

3 0
3 years ago
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