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Dvinal [7]
3 years ago
6

Can someone please help!!!!!

Mathematics
1 answer:
Artist 52 [7]3 years ago
3 0

Answer:

The answer is 10%

Step-by-step explanation:

Let the length of the rectangle is x

Let the width of the rectangle is y

∵ They will increase by 10% each

∴ The new length = x × 110/100 = 1.1x

∴ The new width = y × 110/100 = 1.1y

∵ The old perimeter = 2(x + y) = (2x + 2y)

∴ The new perimeter = 2(1.1x + 1.1y) = 1.1(2x +2y)

∴ The new perimeter is greater than the old perimeter by 1.1 - 1 = 0.1

∴ The percentage of increasing in the perimeter = 0.1 × 100 = 10%

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Solve the system for x and y -3x+8y=103 -5x-8y=-255
Romashka-Z-Leto [24]

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Answer:

  (x, y) = (19, 20)

Step-by-step explanation:

The coefficients of y are opposites, so we can add the two equations to eliminate y.

  (-3x +8y) +(-5x -8y) = (103) +(-255)

  -8x = -152

  x = 19 . . . . . . . divide by -8

Substituting into the first equation, we have ...

  -3(19) +8y = 103

  8y = 160

  y = 20

The solution is (x, y) = (19, 20).

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2 years ago
My dad uses a reusable cup every time every time he goes to the coffee shop the coffee shop takes $0.10 off each coffee every ti
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Answer:

there are 7 days in a week x 0.10 cents saved each day on the week = 0.70 cents every week

Step-by-step explanation:

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3 years ago
Given: The coordinates of triangle PQR are P(0, 0), Q(2a, 0), and R(2b, 2c). Prove: The line containing the midpoints of two sid
dsp73

Answer:

The answer is (a, 0)

Step-by-step explanation:

It's correct.

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3 years ago
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One hundred items are simultaneously put on a life test. Suppose the lifetimes
romanna [79]

Answer:

a) \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b) \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( i-1 )st and the ith failures. Then, the T_{i} are independent with \mathrm{T}_{\mathrm{i}} being exponential with rate \frac{(101-i)}{200} .

Therefore,

a) E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]

=\sum_{i=1}^{5} \frac{200}{101-i}

\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b)

The variance is given by, \mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]

\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

7 0
3 years ago
A unicorn daycare center requires there to be 2 supervisors for every 18 baby unicorns. Write an equation that shows the relatio
Over [174]

Answer:

n = (1/9)u

Step-by-step explanation:

This is a direct variation relationship of the form

n = ku

For n = 2, u = 18.

2 = k * 18

We solve for k.

k = 2/18

k = 1/9

Now we use k in the equation of the relationship.

n = (1/9)u

4 0
2 years ago
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