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sergiy2304 [10]

3 years ago

14

the length of a rectangle is 10 ft more than the width. if the perimeter is 64 feet then what is the length of the rectangle in

feet

1 answer:

Maru [420]3 years ago

8 0

Answer:

use area to solve it

Step-by-step explanation:

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If f(x)=3x+1 then f(a+h)-f(a)

Varvara68 [4.7K]

F(x) = 3x+1
put x = a+h
f(a+h) = 3(a+h) + 1
f(a+h) = 3a + 3h + 1
put x = a
f(a) = 3a + 1
f(a+h)-f(a) = 3a + 3h + 1 -(3a+1)
f(a+h)-f(a) = 3a +3h + 1 - 3a -1
f(a+h)-f(a) = 3h

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3 years ago

A farmer has a 100 ft by 200 ft rectangular field that he wants to increase by 15.5% by cultivating a strip of uniform width aro

Advocard [28]

Answer:

(a) The strip should be 5ft wide

(b) The strip around the outside field is 10ft wide.

Step-by-step explanation:

Given:

Length of the rectangular field, L= 200 ft

width of the rectangular field, w = 100 ft

Area of the rectangular field, A = 200ft x 100ft = 20000 ft^2

let the width of the strip = x

The strip around the outside field = 2x

If the field is increased by 15.5%

New area of the field = 1.155 x 20000 = 23,100 ft^2

The increase in area of the field = 3,100 ft

3,100 = New area of field - old area of the field

3100 = (200 + 2x)(100 + 2x) - 20000

3100 = 20000 + 400x 200x + 4x^2 - 20000

3100 = 600x + 4x^2

Divide through by 4

775 = 150x + x^2

x^2 + 150x - 775 = 0

Factorize

(x + 155)(x-5) = 0

x = 5 ft

The strip should be 5ft wide.

The strip around the outside field = 2 x 5 ft = 10 ft

Thus, the strip around the outside field is 10ft wide.

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3 years ago

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3 years ago

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What are the solutions of this quadratic equation?

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3 years ago

You want to wrap a gift shaped like the regular triangular prism shown. How many

kotegsom [21]

Answer:

$697.6 \ in^2$

Step-by-step explanation:

-Given the dimensions of the box is 8.7 in, 14 in and 10 in.

Let length=14 in, width=10 in and height =8.7 in

#The surface area of a rectangular prism is calculated as:

$Surface \ Area=2(wl+hl+hw)\\\\l=length, w=width, h=height\\\\\therefore Surface \ Area=2(14\times 10+10\times 8.7+14\times 8.7)\\\\=697.6 \ in^2$

Hence, the wrapping paper needed has an area of $697.6 \ in^2$

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3 years ago