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tensa zangetsu [6.8K]
3 years ago
11

Write the equation of the perpendicular bisector of AB if A(-6, -4) and B (2,0).

Mathematics
1 answer:
valkas [14]3 years ago
4 0

Answer:

Step-by-step explanation:

Long way:

Find the midpoint (M) of AB, find the slope(m) of the line AB, then find the equation of the line through M with slope -1/m, which is the perpendicular bisector of AB.

Short way:

The perpendicular bisector of (x₁,y₁) and (x₂,y₂): 2(x₂-x₁)x + 2(y₂-y₁)y = x₂²-x₁²+y₂²-y₁².

Here this is 2(1-(-4))x + 2(6-4)y = 1²-(-4)²+6²-4² which is 10x + 4y = 5.

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Given the system of equations presented here:
Nata [24]
2x + 4y = 14
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2x + 4y = 14
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-14x = -66....as u can see, ur y's cancel out

so ur answer is : 1st answer choice <==

** and just so u know, u could have multiplied the 1st equation by -2, and it would have cancelled out ur x's
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Evaluate each expression<br><br> g5-h3 if g=2 and h=7
vovangra [49]
Look at it this way:
g=2
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Why do the inequality signs stay the same in the last two steps of exercise 1
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In science class, students are mixing water and a citric acid solution. The table shows how much water and acid solution two stu
rusak2 [61]

Answer:

the one with the most parts citric acid

Step-by-step explanation:

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What is the volume of the sphere that has a diameter of 3? use 3.14 for pie​
aev [14]

Hey ! there

Answer:

  • <u>1</u><u>1</u><u>3</u><u>.</u><u>0</u><u>4</u><u> </u><u>unit </u><u>cube</u>

Step-by-step explanation:

In this question we are provided with a sphere <u>having</u><u> </u><u>radius </u><u>3 </u><u>units </u>and <u>value </u><u>of </u><u>π </u><u>is </u><u>3.</u><u>1</u><u>4</u><u> </u><u>.</u><u> </u>And we're asked to find the<u> </u><u>volume</u><u> of</u><u> </u><u>sphere</u><u> </u><u>.</u>

For finding volume of sphere , we need to know its formula . So ,

\qquad \qquad \: \underline{\boxed{ \frak{Volume_{(Sphere)} =  \dfrac{4}{3} \pi r {}^{3} }}}

<u>Where</u><u> </u><u>,</u>

  • π refers to <u>3.</u><u>1</u><u>4</u>

  • r refers to <u>radius</u><u> of</u><u> sphere</u>

<u>Sol</u><u>u</u><u>tion </u><u>:</u><u> </u><u>-</u>

Now , we are substituting value of π and radius in the formula ,

\quad \longrightarrow \qquad \: \dfrac{4}{3}   \times 3.14 \times (3) {}^{3}

Simplifying it ,

\quad \longrightarrow \qquad \: \dfrac{4}{3}  \times 3.14 \times 3 \times 3 \times 3

Cancelling 3 with 3 :

\quad \longrightarrow \qquad \: \dfrac{4}{ \cancel{3}}  \times 3.14 \times 3 \times 3 \times  \cancel{3}

We get ,

\quad \longrightarrow \qquad \:4 \times 3.14 \times 9

Multiplying 4 and 3.14 :

\quad \longrightarrow \qquad \:12.56 \times 9

Multiplying 12.56 and 9 :

\quad \longrightarrow \qquad \:    \pink{\underline{\boxed{\frak{113.04  \: unit \: cube}}}} \quad \bigstar

  • <u>Henceforth</u><u> </u><u>,</u><u> </u><u>volume</u><u> </u><u>of</u><u> </u><u>sphere</u><u> </u><u>having </u><u>radius </u><u>3 </u><u>units </u><u>is </u><em><u>1</u></em><em><u>1</u></em><em><u>3</u></em><em><u> </u></em><em><u>.</u></em><em><u>0</u></em><em><u>4</u></em><em><u> </u></em><em><u>units </u></em><em><u>cube </u></em><em><u>.</u></em>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>

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