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Crazy boy [7]
3 years ago
11

A scale drawing for a rectangular parking lot measures 6.8 cm by 12.3 cm. The scale is 5 cm : 25 m. Find the area of the parking

lot.
Mathematics
1 answer:
Lena [83]3 years ago
8 0
The area is 2091 m squared because if you make a ratio of centimeters over meters and solve for how long each side is which should be 34m and 61.5m. Then, to find the area, use the formula A=lw to find that the area is 2091
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The sides of a rectangle are a number and 4 less than that same number. The perimeter is 56. Find the dimensions of the rectangl
mafiozo [28]
X+x+x-4+x-4=56
4x-8=56
4x=64
x=16

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3 years ago
My stoopid ahh needs help with something else someone HELP
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B because you don’t know exactly how many times she’s gonna grab the marbles
7 0
3 years ago
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Which equation is parallel to the line LaTeX: y=\frac{1}{2}x+3y = 1 2 x + 3and passes through the point (10, -5)?
sukhopar [10]

Answer:

Equation\ of\ line:\ y=\frac{1}{2}x-10

Step-by-step explanation:

Let\ the\ required\ equation\ is\ y=mx+c\\\\where\ m\ is\ the\ slope\ of\ the\ equation\ and\ c\ is\ y-intercept\\\\It\ is\ parallel\ to\ the\ equation\ y=\frac{1}{2}x+3\\\\Hence\ slope\ of\ these\ two\ lines\ will\ be\ same.\\\\Slope\ of\ y=\frac{1}{2}x+3\ is\ \frac{1}{2}\\\\Hence\ slope\ of\ y=mx+c\ is\ \frac{1}{2}\Rightarrow m=\frac{1}{2}\\\\Equation:y=\frac{1}{2}x+c\\\\Line\ passes\ through\ (10,-5).\ Hence\ this\ point\ satisfies\ the\ equation\ of\ line.\\\\-5=\frac{1}{2}\times 10+c

-5=-5+c\\\\c=-10

Equation\ of\ line:\ y=\frac{1}{2}x-10

8 0
3 years ago
A triangle has sides with lengths of 3 feet, 6 feet, and 8 feet is it a right triangle?
Zielflug [23.3K]

Answer:

Solution,\\Longest~side(h) = 8ft.\\Perpendicular(p) = 3ft.\\Base(b) = 6ft.\\Now, \\h^2 = (8)^2 = 64sq.ft.\\p^2+b^2 = 3^2+6^2 = 9+36 = 45sq.ft.\\Since, h^2\neq p^2+b^2, ~the~triangle~is~not~a~right~triangle.

8 0
3 years ago
A person wants to create a vegetable garden and keeps the rabbits out by enclosing it with 100 feet of fencing. The area of the
alekssr [168]

Answer:

The garden can not have area of 700ft^2

Step-by-step explanation:

We are given equation for area as

A(w)=w\times (50-w)

where

w is the width (in feet) of the garden

now, we are given area =700 ft^2

so, we can set area =700

and then we can solve for w

w\times (50-w)=700

50w-w^2=700

-w^2+50w-700=0

now, we can use quadratic formula

ax^2+bx+c=0

w=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

we can compare and find a,b, and c

a=-1 , b=50 , c=-700

now, we can plug values

w=\frac{-50\pm \sqrt{50^2-4\left(-1\right)\left(-700\right)}}{2\left(-1\right)}

w=25-5\sqrt{3}i,\:w=25+5\sqrt{3}i

We can see that values of w is not real

so, w does not exists when area =700

so, The garden can not have area of 700ft^2


4 0
3 years ago
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