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ruslelena [56]
3 years ago
5

Find an expression which represents the difference when (4x – 2y) is subtracted

Mathematics
1 answer:
MatroZZZ [7]3 years ago
7 0

Answer:

-3y

Step-by-step explanation:

"(4x – 2y) subtracted from (4x – 5y)"

(4x – 5y) - (4x – 2y) =

Drop the first set of parentheses since it is unnecessary.

= 4x – 5y - (4x – 2y)

Distribute the negative sign to the left of the parentheses by changing each sign inside the parentheses.

= 4x – 5y - 4x + 2y

Combine like terms.

= 4x - 4x - 5y + 2y

= 0 - 3y

= -3y

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Chris is 6 feet tall. To find the height of a tree, he measures his shadow and the tree's
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3 years ago
A rectangle has a length of 45 inches less than 4 times its width. if the area of the rectangle is 3325 square inches, find the
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The length of the rectangle is based on its width, so let's call width w and put the length in terms of the width. We are told that the width is 45 less than 4 times the width, so the length is 4w - 45. Area is found by multiplying length times width and we are given the area as 3325. So we will set up length times width and solve for w, which we will then use to solve for l. 3325 = (4w - 45)(w). Multiplying out we have 3325=4w^2-45w. Move the constant over by subtraction and then we will have a quadratic that can be factored to solve for w. 4w^2-45w-3325=0. We would put that through the quadratic formula to solve for w. When we do that we get that w = 35 and w = -23.75. The 2 things in math that will never EVER be negative is time and distance/length, so -23.75 is out. That means that the width is 35. The length is 4(35) - 45 which is 95. The dimensions of your rectangle are length is 95 and width is 35. There you go!

8 0
3 years ago
slader (a) Find parametric equations for the line through (4, 1, 8) that is perpendicular to the plane x − y + 4z = 2. (Use the
Pavel [41]

Answer:

(x(t), y(t), z(t)) = (4 + t, 1 - t, 8 + 4t)

xy - plane    (x, y, z) = (2, -1, 0)

yz - plane    (x, y, z) = (0, 5, -8)

xz - plane     (x, y, z) = (5, 0, 12)

Step-by-step explanation:

The given point (x, y ,z) = (4, 1, 8)

The plane x -y + 4z = 2

Normal vector (n) = < 1, -1, 4 >

The equation of line through point (4, 1, 8) and the plane is:

(x(t), y(t), z(t)) = (4, 1, 8) + t(1, -1, 4)

(x(t), y(t), z(t)) = (4 + t, 1 - t, 8 + 4t)

Any point on the line P(x, y, z) = ( 4 + t, 1 - t, 8 + 4t)

xy-Pane ⇒ z = 0

8 + 4t = 0

4t = - 8

t = -8/4

t = -2

∴

(x, y, z) = (4 - 2, 1 - 2, 8 + 4(-2))

(x, y, z) = (2, -1, 0)

yz-plane ⇒ x = 0

4 + t = 0

t = -4

∴

(x, y, z) = (4 + (-4) , 1-(-4), 8 + 4(-4)

(x, y, z) = (0, 5, -8)

xz-plane ⇒ y = 0

1 - t = 0

-t = -1

t = 1

∴

(x, y, z) = ( 4 + 1, 1 - 1, 8 + 4(1) )

(x, y, z) = (5, 0, 12)

6 0
3 years ago
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