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Sliva [168]
3 years ago
15

Write an equation for the line through (4,-1) and (6,9).​

Mathematics
2 answers:
artcher [175]3 years ago
3 0

Answer:

y = 5x - 21

Hope this helps!

ozzi3 years ago
3 0
Y=5x-21 that’s the answer
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Kira solved the equation in the following way. 7 2 − 2x = 11 2 7 2 − 7 2 − 2x = 11 2 − 7 2 −2x ÷ −2 = 4 2 ÷ − 2 x = −1 Describe
vladimir2022 [97]

Answer:

the first step is to use the

✔ Subtraction

property of equality to combine the constant terms.

The second step is to use the

✔ Division

property of equality to isolate the variable

Step-by-step explanation:

edge 20 20

7 0
3 years ago
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Can someone help me please?
lakkis [162]
Choice A is correct
BC=DC
the two triangles share one angel C
the two triangles each have a right angle
so they are congruent according to Angle-Side_Side Theorem 
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3 years ago
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Please Help,This is hard. If you could do it step by step that would be Great :)​
Anna11 [10]

A. Bc a negative times a negative would be positive

C. Bc it would equal 0 and zero is not negative.

1*-7=-7 it wants a negative value so times it by one keeps it negative and same with if it is Multiplied by 7

8 0
3 years ago
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Can anyone pls help me in dis
Charra [1.4K]

Step-by-step explanation:

The coordinates of point A (2,8) and C (5,2)

The measure of segment AC

\sqrt{ {(x2 - x1)}^{2}  +  {(y2 - y1)}^{2}  }

\sqrt{ {(5 - 2)}^{2}  +  {(2 - 8)}^{2} }

\sqrt{ {(3)}^{2} +  {( - 6)}^{2}  }

\sqrt{9 + 36}

\sqrt{45}

6.7 \: units

8 0
3 years ago
PLZ HELP 70 POINTS!!!!!
marin [14]

5^(x+7)=(1/625)^(2x-13)


We move all terms to the left:


5^(x+7)-((1/625)^(2x-13))=0



Domain of the equation: 625)^(2x-13))!=0


x∈R



We add all the numbers together, and all the variables


5^(x+7)-((+1/625)^(2x-13))=0


We multiply all the terms by the denominator



(5^(x+7))*625)^(2x+1-13))-((=0


We add all the numbers together, and all the variables


(5^(x+7))*625)^(2x-12))-((=0


We add all the numbers together, and all the variables


(5^(x+7))*625)^(2x=0

not sure if this is right :/


8 0
3 years ago
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