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3 years ago

4(x-11)^2=25

1 answer:

8 0

Expand the square: 4(x-11)^ =25
4(x-11)(x-11)=25

Distribute : 4(x-11)(x-11)=25
4(x(x-11)-11(x-11))=25

Distribute: 4(x(x-11)-11(x-11))=25
4(x^-11x-11(x-11))=25

Solution: x=17/2
x=27/2

hope that helps

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Two mechanics worked on a car. The first mechanic worked for

sergij07 [2.7K]

76.25 is the answer because 1525 divided by 20 equals 76.25

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3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago

Find the value of x and y in the following figure where ABCD is a parallelogram<br><br>

asambeis [7]

Answer:

x = 3

y = 2

Step-by-step explanation:

Diagonals of a parallelogram bisect each other into two equal segments. Therefore:

3x - 1 = 2(x + 1)

Solve for x

3x - 1 = 2x + 2

Collect like terms

3x - 2x = 1 + 2

x = 3

Also:

5y + 1 = 6y - 1

Collect like terms

5y - 6y = -1 - 1

-y = -2

Divide both sides by -1

y = -2/-1

y = 2

5 0

3 years ago

PLEASE HELP ME WITH BOTH MATH QUESTIONS ASAP!!!!!!!!!

maxonik [38]

Answer:

the one on top= c= -12

the bottom one= f= -15

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A fruit company delivers its fruit in two types of boxes: large and small. a delivery of 5 large boxes and 3 small boxes has a t

Hoochie [10]

Let's let the weight of a large box be L, and the weight of a small box be S.

We know that 5 large boxes and 3 small boxes is 120kg, so:
5L + 3S = 120

We also know that 7 large boxes and 9 small boxes is 234kg, so:
7L + 9S = 234

You can multiply the first equation by 3 to get:
15L + 9S = 360

See how now both equations have 9S? We can now subtract one from the other:
(15L+9S) - (7L+9S) = 360-234
8L = 126
L = 15.75

Now sub this value back into an equation:
(5x15.75) + 3S = 120
3S = 41.25
S = 13.75

Double check these values
(7x15.75) + (9x13.75)
= 110.25 + 123.75
=234, which is consistent with above.

So a large box is 15.75kg, and a small box is 13.75kg.

Hope this helped

5 0

3 years ago