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4 years ago

I used the pythagorean theorem. Need a second opinion

Mathematics

1 answer:

uysha [10]4 years ago

7 0

Answer:

Step-by-step explanation:

EF^2=18^2-12^2

EF^2=324-144

EF^2=180

EF=root 180

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A+ -12= 45 6th grade

Dmitry_Shevchenko [17]

A = 57
45 + 12 = 57
57 - (-12) = 45

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3 years ago

A , B, C , D probability

BlackZzzverrR [31]

Answer:

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3 years ago

Solve 6 + 5 √ 2 4 9 − 2 x = 7

NISA [10]

$6+5\sqrt{249}-2x=7 \\-2x=7-6-5\sqrt{249} \\-2x\approx-77.9 \\x\approx\frac{-77.9}{2}\approx38.95$

Hope this helps.

7 0

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 x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

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

Whats the radius and center of (x+4)^2+(y-2)^2=9

Fiesta28 [93]

I believe the center would be (-4,2) while the radius should be 3.

8 0

3 years ago