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PilotLPTM [1.2K]
3 years ago
11

URGENT - Use 1.7 as the value of 3. What is the area of equilateral triangle ABC? sq mi

Mathematics
1 answer:
Alexxandr [17]3 years ago
6 0
Hope this helps Use 1.7 as the value<span> of </span>square<span> root of </span>3<span> what is the are od </span>equilateral<span> of </span>triangle abc<span>? The base is 12 </span>miles<span> </span>
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What is solution set for the quadratic equation ×^2 + 10×+25=0​
Roman55 [17]

Answer:

x=-5

Step-by-step explanation:

x squared+10x+25 factors into (x+5)(x+5). Set x equal to 0 and you get x=-5. To check, plug in negative 5 and you get 25+ -50 + 25=0 which is true.

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3 years ago
How do you round to the greatest place value
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3 years ago
Read 2 more answers
Jacob sells homemade skateboard decks for a profit of $27 per deck. He is considering switching to a new type of material that w
BARSIC [14]

Answer:

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Step-by-step explanation:

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3 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
3 years ago
What are the solutions to x2 + 8x + 7 = 0?
nikitadnepr [17]

Answer:

x= -1 and x= -7

hope it helps.

5 0
3 years ago
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