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3 years ago

Find the center and radius of the circle and sketch its graph X^2+y^2=9

Mathematics

1 answer:

Vaselesa [24]3 years ago

6 0

Answer:

Centre (0,0)

Radius √9 = 3

I've attached a graph if the circle with it

Answered by GAUTHMATH

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Find the equation of the linear function represented by the table below<br> in slope-intercept form.

Arturiano [62]

Answer:

Equation of a line, y = mx+c

Slope (m)= (Y2-Y1)/(X2-X1), let Y2=7, Y1=3 , X2=2 and X1=1

m=(7-3)/(2-1)

= 4/1

= 4

Picking any value of y and its corresponding x including m, put them into the equation of the line

y=mx+c

3 =4(1)+c

3=4+c

3-4=c

-1=c

y=mx+c

y=4x+(-1)

y=4x-1

8 0

2 years ago

M24 = 115°<br> mZ1 = (2x)<br> m22 = (x + 28)

velikii [3]

Answer:

n the figure, m 3 = 75 and mZ10 = 115. ... CY 337] . 2x +13269. 2x =50. X=28. Find mZ1 in each figure. Exoor. 10. تا ... m24 = 75 24 and 22 are vertical angles.

Missing: m22 = ‎| Must include: m22 =

Step-by-step explanation:

8 0

2 years ago

Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]

Anit [1.1K]

Answer:

The answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

Step-by-step explanation:

$\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}$

Solve the L.H.S part:

$\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]$

After calculating the L.H.S part compare the value with R.H.S:

$\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\$

$\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\$

In equation (i) multiply by 3 and subtract by equation (iii):

$\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to c= - \frac{1}{2}$

put the value of c in equation (i):

$\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\$

In equation (ii) multiply by 3 then subtract by equation (iv):

$\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\$

put the value of d in equation (iv):

$\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2$

The final answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

4 0

3 years ago

Brainiest for An Actuate answer

const2013 [10]

Answer:

$cos(x)=\frac{8}{17}$