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3 years ago

Write the equation of this circle in standard form

Mathematics

1 answer:

givi [52]3 years ago

3 0

Answer:

Step-by-step explanation:

because if you solve the problem it is equal to c

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https://app31.studyisland.com/userFiles/nc6line1.gif Which of the following are alternate exterior angles? A. ∠B and ∠G B. ∠E an

tatuchka [14]

B. ∠E and ∠H are alternate exterior angles

7 0

3 years ago

QUICK 35 POINTS PLUS BRANLIEST. PLEASE ANSWER ALL THE QUESTIONS PROPERLY AND EXPLAIN HOW YOU GOT THEM. TRY TO ANSWER THEM BY TOD

Veronika [31]

About 127.3(square root of 16,200, or 90√2) Look at this problem like a right triangle. Each leg is 90 feet, so the hypotenuse is the square root of 90^2 + 90^2
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7 0

3 years ago

Read 2 more answers

13,22,31,... Find the 31st term. Find the 31st term.

Galina-37 [17]

Answer: 283

Step-by-step explanation:

To do this, it is helpful to get an equation you can use to solve any term.

This equation is:

$13 + 9(n-1)$

So simply plug in 31 for n to get

$13+9(31-1)$

$=13+270$

$=283$

5 0

2 years ago

Read 2 more answers

Which expression is equivalent to -9x-1y-9/-15x5y-3?

FrozenT [24]

Answer: -9x-1y-9/

Step-by-step explanation:

8 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