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hjlf
3 years ago
10

Lena is going to cover her entire garden with topsoil. The garden is 36 feet long by 22 feet wide. A bag of topsoil costs $3 and

covers 300 square feet of area.
How much will the topsoil for Lena's garden cost?
Mathematics
1 answer:
OLEGan [10]3 years ago
7 0
$9.00 because, 36*22=792 Feet.
The reason why it costs $9.00 is because you can't just buy half of the bag, you have to buy the whole thing in this equation.

So, I hope this helped you with this equation because that was really easy.
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Can someone please help me
mrs_skeptik [129]

Step-by-step explanation:

5x=4x+12(corresponding angle)

5x-4x=12

x=12

if it is correct answer then please follow me

6 0
2 years ago
A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea
Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let <em>x</em> be the width and <em>y </em>the length of the rectangular field.

Let <em>C </em>the total cost of the rectangular field.

The side made of heavy duty material of length of <em>x </em>costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are <em>x, y, y</em>.  Thus

C=4x+4y+4y+16x\\C=20x+8y

We know that the total area of rectangular field should be 2250 square yards,

x\cdot y=2250

We can say that y=\frac{2250}{x}

Substituting into the total cost of the rectangular field, we get

C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}

We have to figure out where the function is increasing and decreasing. Differentiating,

\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}

Next, we find the critical points of the derivative

20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30

Because the length is always positive the only point we take is x=30. We thus test the intervals (0, 30) and (30, \infty)

C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0

we see that total cost function is decreasing on (0, 30) and increasing on (30, \infty). Therefore, the minimum is attained at x=30, so the minimal cost is

C(30)=20(30)+\frac{18000}{30}\\C(30)=1200

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0
2 years ago
8(3x – 2) ≤ 2(7x + 12)?
leva [86]

Answer: x≤ 4

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
a nut mixture of peanuts and macadamia nuts at a small fair is $1.00 per pound of peanuts and $3.26 per pound of macadamia nuts.
Studentka2010 [4]
This answer seems really hard but just don't overthink it. You pretty much don't have to do any math. It took me forever to do this because I was thinking so hard but i re-read it and the answer was right infront of me.

1.00p + 3.26n = $248.52
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3 years ago
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morpeh [17]
The answer is 15.50 your welcome
8 0
3 years ago
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