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sveta [45]
3 years ago
8

Mrs. Potts is putting a fence around her rectangular garden the length of the garden is 8 yards the width of the garden is 4 yar

ds how many feet of fencing does Mrs. Pott need
Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

Mrs. Potts needs 72 feet of fencing.

Step-by-step explanation:

You find the perimeter of the rectangular garden, which is 24 yards. You then convert the yards to feet. There are 3 feet in a yard. You multiply 24 by 3 to get your final answer of 72 feet.

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Parametric equation of the line d: 4x-3y+17=0
zmey [24]

Answer:

x = t/4 and y = (t+17)/3

Step-by-step explanation:

4x-3y+17=0

or, 4x=3y-17

let, 4x = t

so, x = t/4

and 3y-17 = t

or, 3y = t+17

or, y = (t+17)/3

Answered by GAUTHMATH

6 0
3 years ago
At what x value does the function given below have a hole?<br><br> f(x)=x+3/x2−9
S_A_V [24]

Answer:

hole at x=-3

Step-by-step explanation:

The hole is the discontinuity that exists after the fraction reduces. (Still doesn't exist for original of course)

The discontinuities for this expression is when the bottom is 0. x^2-9=0 when x=3 or x=-3 since squaring either and then subtracting 9 would lead to 0.

So anyways we have (x+3)/(x^2-9)

= (x+3)/((x-3)(x+3))

Now this equals 1/(x-3) with a hole at x=-3 since the x+3 factor was "cancelled" from the denominator.

4 0
3 years ago
The value of 8 in 56982 is how many times larger than the value of 8 in 156408
Len [333]
The value of 8 in '56982' is 80 whereas the value of 8 in '156408' is just 8 so the value of the 8 in 56982 is x10 larger than the value of 8 in 156408.
Hope this helps x
3 0
3 years ago
Read 2 more answers
What is the interquartile range of the data set below? Growth in feet of oak trees: 68,80,73,90,120,94,76,112,101,94,72 (1) 22  
sdas [7]
<span>68,80,73,90,120,94,76,112,101,94,72 --->
68, 72, 73, 76, 80, 90, 94, 94, 101, 112, 120

Median = 90
Lower Median = 73
Upper Median = 101

IQR = 101 - 73
IQR = 28</span>
8 0
3 years ago
A function is given. Determine the average rate of change of the function between the given values of the variable: f(x)= 4x^2;
nevsk [136]
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\bf f(x)= 4x^2  \qquad &#10;\begin{cases}&#10;x_1=3\\&#10;x_2=3+h&#10;\end{cases}\implies \cfrac{f(3+h)-f(3)}{(3+h)~-~(3)}&#10;\\\\\\&#10;\cfrac{[4(3+h)^2]~~-~~[4(3)^2]}{\underline{3}+h-\underline{3}}\implies \cfrac{4(3^2+6h+h^2)~~-~~4(9)}{h}&#10;\\\\\\&#10;\cfrac{4(9+6h+h^2)~~-~~36}{h}\implies \cfrac{\underline{36}+24h+4h^2~~\underline{-~~36}}{h}&#10;\\\\\\&#10;\cfrac{24h+4h^2}{h}\implies \cfrac{\underline{h}(24+4h)}{\underline{h}}\implies 24+4h
8 0
3 years ago
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