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RSB [31]
3 years ago
5

Plss help lots of point plss give the step by step solution to get the answer as well!!!!​

Mathematics
1 answer:
just olya [345]3 years ago
4 0

Answer:(2,0)

Step-by-step explanation:

5x+2y=10

Verification of points

5(2)+2(0)=10

10=10

LHS=RHS

THUS solution is (2,0)

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In triangle STU s=9 cm t=15 and m
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3 years ago
T is in the interior of angle PQR. Find each of the following. Hint: Draw each angle.
balandron [24]

Answer:

13 degrees

Step-by-step explanation:

mPQR=25 degrees , mRQT=11 degrees.

mPQT=PQR-RQT

MPQT=25-11=13 degrees

7 0
3 years ago
Find a point on the y-axis that is equidistant from the points (8, −8) and (1, 1).
artcher [175]

\text{We know that any point on the y-axis is of the form (0, y),}\\
\text{because on y-axis we have x=0.}\\
\\
\text{So let (0, y) be the point on the y-axis that is equidistance from}\\
\text{the points }(8,-8) \text{ and }(1,1)\\
\\
\text{so using the distance formula, we have}\\
\\
\text{Distance between (0, y) and }(8,-8)=\text{Distance between (0,y) and }(1,1)\\
\\
\Rightarrow \sqrt{(0-8)^2+(y-(-8))^2}=\sqrt{(0-1)^2+(y-1)^2}

\Rightarrow \sqrt{64+(y+8)^2}=\sqrt{1+(y-1)^2}\\
\\
\text{Squaring both sides, we get}\\
\\
(\sqrt{64+(y+8)^2})^2=(\sqrt{1+(y-1)^2})^2\\
\\
\Rightarrow 64+(y+8)^2=1+(y-1)^2\\
\\
\Rightarrow 64+(y^2+16y+64)=1+(y^2-2y+1)\\
\\
\Rightarrow y^2+16y+128=y^2-2y+2\\
\\
\Rightarrow y^2+16y-y^2+2y=2-128\\
\\
\Rightarrow 18y=-126\\
\\
\Rightarrow y=\frac{-126}{18}\\
\\
\Rightarrow y=-7

So the point on the y-axis that is equidistant from the points (8,-8) and (1,1) is: (0,-7)

3 0
4 years ago
Need to know the answer I don't get it
Svetllana [295]
The first two are alike because you have to add 3 to each side for everything to come out even.
4 0
4 years ago
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salantis [7]

Answer:

12 hour                24 hour

08:05pm               20:05pm

12:01am                  00:01am              

Step-by-step explanation:

7 0
3 years ago
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