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3 years ago

Passes through (-5,9) and (1,3)

Mathematics

1 answer:

Goshia [24]3 years ago

6 0

Answer:

m=-1

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(3-9)/(1-(-5))

m=-6/(1+5)

m=-6/6

m=-1

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Multiply the polynomials 3(x+7) (show work pls)

Evgen [1.6K]

Answer:

3x + 21

Step-by-step explanation:

(3)(x+7)

Now, we distribute the 3 in each term of (x+7)

So, 3*x = 3x and 3*7 = 21.

So our resulting term would be 3x+21.

8 0

3 years ago

Write the equation of the circle below to complete the square. Also determine the area of the circle.

diamong [38]

Answer: $(x-7)^2+(y+2)^2=11^2$ ; $A=121\pi$

Step-by-step explanation:

I have never done a problem like this, so this is my first time attempting it.

The standard form for the equation of a circle is:

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

Begin by grouping the 'x' terms and 'y' terms:

$(x^2-14x)+(y^2+4y-68)=0$

Complete the square for the second polynomial:

$(\frac{b}{2} )^2=(\frac{4}{2} )^2=4$

$(x^2-14x)+(y^2+4y+4-4-68)=0$

$(x^2-14x)+(y^2+4y+4)-4-68=0$

$(x^2-14x)+(y^2+4y+4)-72=0$

$(x^2-14x)+(y+2)^2-72=0$

Do the exact same process with the other polynomial:

$(x^2-14x-72)+(y+2)^2=0$

Complete the square:

$(\frac{14}{2} )^2=49$

$(x^2-14x+49-49-72)+(y+2)^2=0$

$(x^2-14x+49)-49-72+(y+2)^2=0$

$(x^2-14x+49)-121+(y+2)^2=0$

$(x-7)^2-121+(y+2)^2=0$

Bring the 121 to the other side to get:

$(x-7)^2+(y+2)^2=121$

$(x-7)^2+(y+2)^2=11^2$

The above expression is the equation of our circle in standard form. It is centered at the coordinates (7, -2) and has a radius of 11 units. The area of a circle if given by:

$A=\pi r^2=121\pi$

8 0

1 year ago

Line r is parallel to the graph of 2x – 3y = –18. The y-intercept of line ris

KATRIN_1 [288]

Answer:

3 or (3,0)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Solve the system of equations using matrices. Use the Gauss- Jordan elimination method And find a solution set

KATRIN_1 [288]

$\begin{gathered} x+y+z=4 \\ x-y-z=0 \\ x-y+z=8 \\ \text{The system using matrix is} \\ \begin{bmatrix}{1} & {1} & {1}, & {4} \\ {1} & {-1} & {-1,} & {0} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix}\rightarrow F2=F2-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix} \\ \rightarrow F3=F3-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F2=-\frac{1}{2}F2 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F3=F3+2F2=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {2,} & {8}{}{}\end{bmatrix}\rightarrow F3=\frac{1}{2}F3 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F2=F2-F3=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \rightarrow F1=F1-F3=\begin{bmatrix}{1} & {1} & {0}, & {0} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F1=F1-F2 \\ =\begin{bmatrix}{1} & {0} & {0}, & {2} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \text{Therefore, the solution is }x=2,\text{ y=-2 and z=4} \end{gathered}$

6 0

1 year ago

Find the equation of the linear function represented by the table below in slope-

pav-90 [236]

Answer: $y=-3x-2$