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Tju [1.3M]
3 years ago
9

Need help with this problem, doesn't make sense to me.

Mathematics
1 answer:
skad [1K]3 years ago
4 0

Answer:

<u>A = x² - 5x - 8</u>

Step-by-step explanation:

<u>Finding x</u>

  • Area (deck) = Area (Rectangle) - [Area (bench) + Area (planter)]
  • A = x(x - 4) - [ 2(6) + 1/2(2)(x - 4) ]
  • A = x(x - 4) - [12 + x - 4]
  • A = x² - 4x - 8 - x
  • <u>A = x² - 5x - 8</u>
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I will mark you brainiest if you can answer this
Anettt [7]

Answer:

slope- 40

y-intercept- 10

slope-intercept form- y=40x+10

Step-by-step explanation:

hope this helps :)

6 0
3 years ago
Two buildings on opposites sides of a highway are 3x^3- x^2 + 7x +100 feet apart. One building is 2x^2 + 7x feet from the highwa
iVinArrow [24]

Given:

Distance between two buildings = 3x^3- x^2 + 7x +100 feet apart.

Distance between highway and one building = 2x^2 + 7x feet.

Distance between highway and second building = x^3 + 2x^2 - 18 feet.

To find:

The standard form of the polynomial representing the width of the highway between the two building.

Solution:

We know that,

Width of the highway = Distance between two buildings - Distance of both buildings from highway.

Using the above formula, we get the polynomial for width (W) of the highway.

W=3x^3- x^2 + 7x +100-(2x^2 + 7x)-(x^3 + 2x^2 - 18)

W=3x^3- x^2 + 7x +100-2x^2-7x-x^3 -2x^2+18

Combining like terms, we get

W=(3x^3-x^3)+(- x^2 -2x^2-2x^2)+ (7x -7x)+(100 +18)

W=2x^3-5x^2+0+118

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Therefore, the width point highway is 2x^3-5x^2+118.

8 0
3 years ago
Can someone please help me with question 3
Anastasy [175]

Answer:

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Now plug in 4.6 for t you get:

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