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mariarad [96]
3 years ago
14

Pls help!!! Will mark brainliest!! links will be reported.

Mathematics
1 answer:
Ad libitum [116K]3 years ago
3 0

I hope this is the right answer and if it's not then my apologies.

And sorry it took so long.

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Find the point-slope equation for the line
aliya0001 [1]

Answer:

y-30=2(x-2)

Step-by-step explanation:

m=y2-y1/x2-x1

-28-2=(-30)

15-30=(-15)

-30/(-15)=2

m=2

hope this helps =3

sorry if im wrong

6 0
3 years ago
Please help with 17 and/or 18
Marizza181 [45]

Answer:

Slope Intercept Form: y=3x+10

Slope: 3

y: 10

Step-by-step explanation:

y-3x=10    Add 3x to both sides. The y has to be isolated.

So it is y=10+3x     So we have to reorder it so y=3x+10

8 0
3 years ago
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If 2x +y=8, x+2y=4, find the value of x-y.<br> Please, help me
kifflom [539]

y = 8 -2x

x + 2 (8 - 2x) = 1

x + 16 - 4x = 1

x - 4x = -15

-3x = -15

x = 5

If x = 5, then we can substitute in...

5 + 2y = 1

2y = -4

y = -2

Now let us check that by substituting those values in to the equation

2x + y = 8

2 (5) + (-2) = 8

That is correct so we have the right values for x and y/

8 0
3 years ago
Pleaseeeee neeeddd helppp asappp
Rufina [12.5K]

ANSWER:

Segment TS is a Chord

If AR = 14 , then QS = 14 ×2 = 28

∆ATS is an Isosceles Triangle

<em><u>HOPE</u></em><em><u> </u></em><em><u>THIS</u></em><em><u> HELP</u></em><em><u> </u></em><em><u>YOU </u></em><em><u>,</u></em><em><u> FOR</u></em><em><u> ANY</u></em><em><u> QUESTION</u></em><em><u> PLEASE</u></em><em><u> ASK</u></em><em><u> </u></em><em><u>ME</u></em>

7 0
2 years ago
Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

7 0
4 years ago
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