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4 years ago

Equation for given circle. A(4,7) is at center top of circle and B(7,4) is at center right hand side

Mathematics

1 answer:

oee [108]4 years ago

5 0

Note:
The equation for a circle with center at (h,k) and radius r is
(x-h)² + (y-k)² = r²

For the specified circle,
h=4, h=7.

From the diagram shown below, the radius, r, is determined from the Pythagorean Theorem as
r² = 3² + 3² = 18

Therefore the equation for the circle is
(x - 4)² + (y - 7)² = 18

Answer: (x - 4)² + (y - 7)² = 18

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A triangular city lot bounded by three streets has a length of 300 feet on one street, 250 feet on the second, and 420 feet on t

Butoxors [25]

Consider using the Law of Cosines, because the lengths of three sides are given and the largest angle is the one to be approximated. This angle will be opposite the longest side, that is, opposite the 420-foot side.

420^2 = 250^2 + 300^2 - 2(250)(300)cos A.

Then: 176400 = 62500 + 810000 - 150000cos A.

Solving for cos A, we get:

150000cos A = 176400-62500-810000, or -696100

Then:

-696100

cos A = ------------------- = - 4.64. This is not possible, as the range of the cosine

150000 function is [-1,1].

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3 years ago

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Combine the following: a+a+a+b+b+c<br> i really need help <br> help me plzzzzzzzzz

Andrei [34K]

a + a + a is the same as multiplying a by 3 so you would have 3a

b + b is the same as multiplying b by 2 so you would have 2b

Combing everything together you would have the final answer of :

3a + 2b + c

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3 years ago

A pizza place offers 2 different types of crust, 3 different types of cheese, and the option of 5 different toppings. How many d

ipn [44]

if we're going with single topping, single cheese per pizza, the answer is 30.

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3 years ago

Suppose a rectangular pasture is to be constructed using 1 2 linear mile of fencing. The pasture will have one divider parallel

timama [110]

Answer:

$\displaystyle A=\frac{1}{192}$

Step-by-step explanation:

<u>Maximization With Derivatives</u>

Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.

We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.

The two divisions parallel to x have lengths y, thus the fencing will take 4x. The three dividers parallel to y have lengths x, thus the fencing will take 3y.

The amount of fence needed to enclose the external and the internal divisions is

$P=4x+3y$

We know the total fencing is 1/2 miles long, thus

$\displaystyle 4x+3y=\frac{1}{2}$

Solving for x

$\displaystyle x=\frac{\frac{1}{2}-3y}{4}$

The total area of the pasture is

$A=x.y$

Substituting x

$\displaystyle A=\frac{\frac{1}{2}-3y}{4}.y$

$\displaystyle A=\frac{\frac{1}{2}y-3y^2}{4}$

Differentiating with respect to y

$\displaystyle A'=\frac{\frac{1}{2}-6y}{4}$

Equate to 0

$\displaystyle \frac{\frac{1}{2}-6y}{4}=0$

Solving for y

$\displaystyle y=\frac{1}{12}$

And also

$\displaystyle x=\frac{\frac{1}{2}-3\cdot \frac{1}{12}}{4}=\frac{1}{16}$

Compute the second derivative

$\displaystyle A''=-\frac{3}{2}.$

Since it's always negative, the point is a maximum

Thus, the maximum area is

$\displaystyle A=\frac{1}{12}\cdot \frac{1}{16}=\frac{1}{192}$

6 0

3 years ago

Determine what shape is formed for the given coordinates for ABCD, and then find the perimeter and area as an exact value and ro

Helga [31]

Answer:

Part 1) The shape is a trapezoid

Part 2) The perimeter is $25(4+\sqrt{2})\ units$ or approximately $135.4\ units$

Part 3) The area is $937.5\ units^2$

Step-by-step explanation:

step 1

Plot the figure to better understand the problem

we have

A(-28,2),B(-21,-22),C(27,-8),D(-4,9)

using a graphing tool

The shape is a trapezoid

see the attached figure

step 2

Find the perimeter

we know that

The perimeter of the trapezoid is equal to

$P=AB+BC+CD+AD$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

we have

A(-28,2),B(-21,-22)

substitute in the formula

$d=\sqrt{(-22-2)^{2}+(-21+28)^{2}}$

$d=\sqrt{(-24)^{2}+(7)^{2}}$

$d=\sqrt{625}$

$d_A_B=25\ units$

Find the distance BC

we have

B(-21,-22),C(27,-8)

substitute in the formula

$d=\sqrt{(-8+22)^{2}+(27+21)^{2}}$

$d=\sqrt{(14)^{2}+(48)^{2}}$

$d=\sqrt{2,500}$

$d_B_C=50\ units$

Find the distance CD

we have

C(27,-8),D(-4,9)

substitute in the formula

$d=\sqrt{(9+8)^{2}+(-4-27)^{2}}$

$d=\sqrt{(17)^{2}+(-31)^{2}}$

$d=\sqrt{1,250}$

$d_C_D=25\sqrt{2}\ units$

Find the distance AD

we have

A(-28,2),D(-4,9)

substitute in the formula

$d=\sqrt{(9-2)^{2}+(-4+28)^{2}}$

$d=\sqrt{(7)^{2}+(24)^{2}}$

$d=\sqrt{625}$

$d_A_D=25\ units$

Find the perimeter

$P=25+50+25\sqrt{2}+25$

$P=(100+25\sqrt{2})\ units$

simplify

$P=25(4+\sqrt{2})\ units$ ----> exact value

$P=135.4\ units$

therefore

The perimeter is $25(4+\sqrt{2})\ units$ or approximately $135.4\ units$

step 3

Find the area

The area of trapezoid is equal to

$A=\frac{1}{2}[BC+AD]AB$

substitute the given values

$A=\frac{1}{2}[50+25]25=937.5\ units^2$

4 0

3 years ago