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4 years ago

Three vertices of a rectangle are given. Find the coordinates of the fourth vertex.

1 answer:

7 0

So if we imagine a rectangle in our head, we know that there will be coresponding heights(y) and coresponding widths(x) we look at the y values given (x,y) -3,-3,2 the other y value must be 2 the answer is either B or D look at the x values -3,7,7 the x value must be -3 the answer is x=-3 y=2 (-3,2) the answer is D (-3,2)

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Please let me know the correct answer.

NNADVOKAT [17]

Set h to 640 and solve for t:

640 = -490t^2 + 1120t

Subtract 640 from both sides:

-490t^2 + 1120t - 640 = 0

The formula to solve a quadratic equation is:

x = -b -/+ sqrtroot (b^2-4ac)/(2a) where a = -490, b = 1120 and c = -640

Solve:

x = -1120 -/+ sqrtroot (1120^2-4(-490)(-640) )/ 2(-490)

x = 8/7 = 1.1428 = 1.14

Time was 1.14 seconds

4 0

3 years ago

Somebody please help I hate geometry

borishaifa [10]

Answer:

-----

Step-by-step explanation:

mark me as brilliant

7 0

3 years ago

If I have (7x+3) degrees and (8x+1) degrees what would that equal

Sidana [21]

Answer:

180

Step-by-step explanation:

i just did it got it right

8 0

3 years ago

How to do this. It’s hard

Black_prince [1.1K]

I think the answer is y=10

4 0

3 years ago

Read 2 more answers

Which equation represents the circle described?

bazaltina [42]

We have the equation:

$x^2+y^2-8x-6y+24=0$

By arranging this equation in terms of x and y, we have:

$x^2-8x+y^2-6y=-24 \\ \\$

By using the method of completing the square, we have:

$x^2-8x+\mathbf{\left(\frac{8}{2}\right)^2}+y^2-6y+\mathbf{\left(\frac{6}{2}\right)^2}=-24+\mathbf{\left(\frac{8}{2}\right)^2}+\mathbf{\left(\frac{6}{2}\right)^2} \\ \\ x^2-8x+\mathbf{16}+y^2-6y+\mathbf{9}=-24+\mathbf{16}+\mathbf{9} \\ \\ \boxed{(x-4)^2+(x-3)^2=1}$

The center of this circle is:

$(h,k)=(4,3)$

So the equation that fulfills the statement is:

$(x-4)^2+(y-3)^2=2^2$

Finally, the right answer is c)

6 0

3 years ago

Read 2 more answers