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Lostsunrise [7]
2 years ago
15

A rectangular sandbox has a width of 5 feet.The sandbox is 5 times as long as it is wide. What is the perimeter of the sandbox?

Mathematics
1 answer:
AveGali [126]2 years ago
8 0
Wide is 5 x 5 = 25 feet;
P = 2 x width + 2 x wide = 10 + 50 = 60 feet;
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Margot's dad has 7 striped ties and 21 solid ties. What is ratio striped ties to solid ties
Anuta_ua [19.1K]

7 : 21

which can be simplified to 1 : 3 as both sides are divisible by 7

8 0
2 years ago
_ 5x^20-7x^10+15 in quadratic form
Tpy6a [65]

Answer:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -(1/10 (-7 - sqrt(349)))^(1/10) or x = (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10)

Step-by-step explanation:

Solve for x:

-5 x^20 - 7 x^10 + 15 = 0

Substitute y = x^10:

-5 y^2 - 7 y + 15 = 0

Divide both sides by -5:

y^2 + (7 y)/5 - 3 = 0

Add 3 to both sides:

y^2 + (7 y)/5 = 3

Add 49/100 to both sides:

y^2 + (7 y)/5 + 49/100 = 349/100

Write the left hand side as a square:

(y + 7/10)^2 = 349/100

Take the square root of both sides:

y + 7/10 = sqrt(349)/10 or y + 7/10 = -sqrt(349)/10

Subtract 7/10 from both sides:

y = sqrt(349)/10 - 7/10 or y + 7/10 = -sqrt(349)/10

Substitute back for y = x^10:

x^10 = sqrt(349)/10 - 7/10 or y + 7/10 = -sqrt(349)/10

Taking 10^th roots gives (sqrt(349)/10 - 7/10)^(1/10) times the 10^th roots of unity:

x = -(1/10 (sqrt(349) - 7))^(1/10) or x = (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(1/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(1/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(2/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(2/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(3/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(3/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(4/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(4/5) (1/10 (sqrt(349) - 7))^(1/10) or y + 7/10 = -sqrt(349)/10

Subtract 7/10 from both sides:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or y = -7/10 - sqrt(349)/10

Substitute back for y = x^10:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x^10 = -7/10 - sqrt(349)/10

Taking 10^th roots gives (-7/10 - sqrt(349)/10)^(1/10) times the 10^th roots of unity:

Answer:  x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -(1/10 (-7 - sqrt(349)))^(1/10) or x = (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10)

8 0
3 years ago
3. If a circle has a circumference of T units, what is the diameter of the circle?
musickatia [10]

Answer:

Diameter = \frac{T}{\pi}  units

Step-by-step explanation:

There are 2 formulas that we have for circumference of a circle, shown below.

  1. C=2\pi r
  2. C=\pi d

This comes from the fact that twice radius (r) is equal to diameter (d).

Since the question asks for diameter given circumference, we will use #2 formula from above because it relates circumference and diameter.

C = T [given]

d = ???

Let's find by substituting in formula:

C=\pi d\\T = \pi d\\d=\frac{T}{\pi}

Hence the diameter is  \frac{T}{\pi}  units

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siniylev [52]

Answer:

k = 8

Step-by-step explanation:

For the points to be collinear, then the slopes of each pair of points must be equal.

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 2, 2) and (x₂, y₂ ) = (2, 6)

m = \frac{6-2}{2+2} = \frac{4}{4} = 1

Repeat with

(x₁, y₁ ) = (2, 6) and (x₂, y₂ ) = (4, k)

m = \frac{k-6}{4-2} = \frac{k-6}{2}, thus

\frac{k-6}{2} = 1 ( multiply both sides by 2 )

k - 6 = 2 ( add 6 to both sides )

k = 8

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F.

have a great day:)
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