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3 years ago

Help I will be marking brainliest!!!

Mathematics

2 answers:

Effectus [21]3 years ago

7 0

Answer:

102 is the correct answet

ratelena [41]3 years ago

6 0

Answer:

brainliest please

okay so angle O is connected to arc GU. to find angle O you musit find Arc GU

if Arc GO is 48 degrees and arc OU is 107 degrees, then you add those together.

107 +48=155

a circle has a total of 360 degrees

so 360-155= 205

205 is arc GU

205 divided by 2 will give you angle O

O is 102.5 degrees

Step-by-step explanation:

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X+y=3 and x-ý= x1 simultaneous equation

pshichka [43]

Answer:

y=1, x=2

Step-by-step explanation:

x+y=3

x-y=1

Using elimination method

x-x+y--y=3-1

2y= 2

y=1

Substitute y=1 into equation 2

x-1=1

x=1+1

x=2

7 0

3 years ago

Read 2 more answers

Part A picture above

-BARSIC- [3]

Answer:

Part A C = (10,5) Part B C. D'(0,10)

Step-by-step explanation:

Part A

Since c is at the point (2,1) in relation to the origin, we can multiply those distances by our scale factor of 5

(2,1) * 5 = (10,5)

The new point C is going to be (10,5)

Part B

If you dilate with a factor of 5 -- relative to the origin -- you have to multiply the distance from the origin by 5.

In this case, point D is already on the y axis, so it's x value wouldn't be affected. Point D is currently 2 units away from (0,0), so we can multiply 2*5 to get 10 -- our ending point is (0,10)

8 0

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

Thank you for helping me

dedylja [7]

Answer:

thank you for giving me points ;)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please help with three through 12

lord [1]