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netineya [11]
3 years ago
14

Please help it's urgent! ​

Mathematics
1 answer:
aleksandrvk [35]3 years ago
4 0

Answer:

25 x 2 = 50 :D

Step-by-step explanation:

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A line passes through (1, –5) and (–3, 7).
Brilliant_brown [7]
Slope of the line =(7- -5) / (-3-1) = 12 / -4 = -3

a. point slope form is y + 5 = -3(x - 1)  Answer

b  rearrange to slope-intercept form:-
 y = -3x + 3 - 5
y = -3x - 2  Answer
5 0
3 years ago
Which is a graph for the inequality m < -2 ?
aliya0001 [1]

Answer:

B

Step-by-step explanation:

This answer shows values of -2, and less than -2

6 0
3 years ago
Find the smallest value of k such that the LCM of<br> k and 6 is 60
Advocard [28]

Answer: The smallest valuest value for<em> k </em>is 10, such that LCM o<em>f k</em> and 6 is 60.

Step-by-step explanation:

We know that, LCM = Least common multiple.

For example : LACM of 12 and 60  is 60.

If LCM of k and 6 is 60.

i.e. the least common multiple of k and 6 is 60.

Since, 10 x 6 = 60.

The smallest valuest value for<em> k </em>should be 10, such that LCM o<em>f k</em> and 6 is 60.

Hence, the smallest value of k is 10.

5 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}\\\\\\&#10;\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}\\\\\\&#10;\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}\\\\\\&#10;

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
Which transformation will map figure Q onto figure Q’
Ilya [14]

Answer:

11 units i think

Step-by-step explanation:

6 0
3 years ago
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