What is slope when the vertical change is-16 and the horizontal change is 2?

The Garden Task

Igoryamba

If the perimeter of the garden is (14x - 32) ft and has a side length of (x + 2) ft:

the perimeter = 24 ft
the area = 36 sq. ft
if the perimeter of the garden is doubled, the perimeter of the new garden = 48 ft
if the area of the garden is doubled, the area of the new garden = 72 sq. ft

Recall:

A square has equal side lengths
Perimeter of a square = 4(side length)
Area of a square = $(side $ length)^2$

Given:

Perimeter of square (P) = $(14x - 32) $ ft$

Side length (s) = $(x + 2) $ ft$

First, let's find the value of x by creating an equation using the perimeter formula:

Perimeter of a square = 4(side length)

Plug in the values

$(14x - 32) = 4(x + 2)$

Solve for x

$14x - 32 = 4x + 8\\\\14x - 4x = 32 + 8\\\\10x = 40\\\\x = 4$

Find how much fencing would be needed (Perimeter of the fence):

Perimeter of the fence = $(14x - 32) $ ft$

Plug in the value of x

Perimeter of the fence = $14(4) - 32 = 24 $ ft$

Find the area of the garden:

Area of the garden = $(side $ length)^2$

Area = $(x + 2)^2$

Plug in the value of x

Area = $(4 + 2)^2 = 6^2 = 36 $ ft^2$

Find the perimeter if the garden size is doubled:

Perimeter of the new garden = 2 x 24 = 48 ft

Find the area if the garden size is doubled:

Perimeter of the new garden = 2 x 36 = 72 sq. ft

In summary, if the perimeter of the garden is (14x - 32) ft and has a side length of (x + 2) ft:

the perimeter = 24 ft
the area = 36 sq. ft
if the perimeter of the garden is doubled, the perimeter of the new garden = 48 ft
if the area of the garden is doubled, the area of the new garden = 72 sq. ft

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4 0

1 year ago

How can bar diagrams be used to model numerical expressions

vesna_86 [32]

A bar diagram can be used to represent algebraic expressions or to give examples.

5 0

2 years ago

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Given: ∆ABC, m∠C = 90° CB = 8, m∠B = 38º Find the area of a circumscribed circle. Find the area of the inscribed circle.

vitfil [10]

Answer:

Circumscribed circle: Around 80.95

Inscribed circle: Around 3.298

Step-by-step explanation:

Since C is a right angle, when the circle is circumscribed it will be an inscribed angle with a corresponding arc length of 2*90=180 degrees. This means that AB is the diameter of the circle. Since the cosine of an angle in a right triangle is equivalent to the length of the adjacent side divided by the length of the hypotenuse:

$\cos 38= \dfrac{8}{AB} \\\\\\AB=\dfrac{8}{\cos 38}\approx 10.152$

To find the area of the circumscribed circle:

$r=\dfrac{AB}{2}\approx 5.076 \\\\\\A=\pi r^2\approx 80.95$

To find the area of the inscribed circle, you need the length of AC, which you can find with the Pythagorean Theorem:

$AC=\sqrt{10.152^2-8^2}\approx 6.25$

The area of the triangle is:

$A=\dfrac{bh}{2}=\dfrac{8\cdot 6.25}{2}=25$

The semiperimeter of the triangle is:

$\dfrac{10.152+6.25+8}{2}\approx 24.4$

The radius of the circle is therefore $\dfrac{25}{24.4}\approx 1.025$

The area of the inscribed circle then is $\pi\cdot (1.025)^2\approx 3.298$ .

Hope this helps!

6 0

3 years ago

Which prism has a volume of 5 cubic units?

erastova [34]

Answer:

Step-by-step explanation:

5=1 1/4*4*1=1.25*4*1

so 1 and one fourth units by 4 units by 1 unit

6 0

3 years ago

Read 2 more answers

A) Work out the value of 5^2 – 1^5× 2^4 b) Work out the value of 7^2 + 4^3 ÷ 2^5

NemiM [27]

First you have to know BODMAS/BIDMAS (idk if all schools teach it like this) but this is the order in which you complete questions so it’s Brackets, Indices, Division/Multiplication, Addition/Subtraction
a) 5^2 is 5x5 which equals 25 1 to the power of anything is always 1 and 2^4 is 2x2x2x2 which is 16
So 5^2-1^5x2^4 = 25-1x16 which is then 25-16 = 9
b) 7^2+4^3 divided by 2^5 so 7^2 is 7x7 which = 49 then 4^3 means 4x4x4 = 64 and 2^5 is 2x2x2x2x2 which equals 32 so
7^2+4^3 divides by 2^5 is 49+64/32 = 49+2 which is 51

8 0

3 years ago