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blondinia [14]
3 years ago
5

Two sides of a right triangle have the lengths 4 and 5. What is the product of the possible lengths of the third side? Express t

he product as a decimal rounded to the nearest tenth.
Mathematics
1 answer:
marishachu [46]3 years ago
6 0

Answer:

19.2

Step-by-step explanation:

<u>1st Case:</u>

4 and 5 are legs of the right triangle.

Using the pythagorean therom: a^2+b^2=c^2

We can say that 4^2+5^2=x^2

16+25=x^2

41=x^2

x=√41

√41 is about 6.4

x=6.4

<u>2nd Case</u>

5 is the hypotenuse of the right triangle and 4 is the legs.

Using the pythagorean therom: a^2+b^2=c^2

We can say that 4^2+x^2=5^2

16+x^2=25

x^2=9

x=3

<u>Final Step</u>

We need to multiply the two possible lengths for x. So for case 1 the length of x was 6.4 and for case two the length was 3. 6.4*3=19.2

Anwser: <u>19.2</u>

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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:

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2 years ago
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Answer:

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Step-by-step explanation:

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I think the answer is K
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