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steposvetlana [31]
3 years ago
14

Another question please help ​

Mathematics
1 answer:
finlep [7]3 years ago
5 0

As we know, The slopes of parallel lines are equal.

We are given the lines

\tt \: y = m_{1} x + c \\  \sf \: and \\  \tt \: y =  m_{2}x + c

Their slopes are m1 and m2, So their relation is

\bf \: m_{1} =  m_{2}

  • <em>Thus, Option A is correct!!~</em>
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AC. DF, and GI are parallel. Use the figure to complete the proportion.<br> AD/DG = ?/EH
ivanzaharov [21]

Answer:

BE.

Step-by-step explanation:

If AD is to DG, then BE must be to EH, and CF must be to FI.

3 0
3 years ago
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What is the equation of the quadratic graph with a focus of (1 ,3) and a directrix of y=1?
zimovet [89]
Check the picture below.

now, we know the directrix is at y = 1, and the focus point is at 1,3, well, notice the picture, the distance between those fellows is just 2 units.

the vertex is half-way between those fellows, therefore, the vertex will be at 1,2.

the distance "p", from the vertex to either the directrix or focus, is really just 1 unit.  Since the focus point is above the directrix, is a vertical parabola, and it opens upwards, like in the picture, and since it opens up, the "p" value is positive, or +1.

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4 0
3 years ago
Show steps to solve <br><br>2-16.5=4-7.5​
ASHA 777 [7]

Answer:

11=0

or

-11=0

keep in mind that you get these answers by completely simplifying the equation, but they are both false

Step-by-step explanation:

step 1: simplify both sides

2-16.5 simplifies to -14.5

4-7.5 simplifies to -3.5

so you get -3.5 = -14.5

step 2: add one side to the other to get a zero on one side

add 3.5 on both sides to get 0=-11

or

add 14.5 on both sides to get 11=0

step 3: evaluate the statement

since both -11 and 11 are not equal to 0, this statement is false

7 0
3 years ago
Janice ran mile. She walked mile. How far did she run and walk?​
Vera_Pavlovna [14]

Answer:

two miles

Step-by-step explanation:

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PLEASEE HELP WILL GIVE 100 POINTS AND WILL MARK BRAINLY!
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I’m pretty sure it’s b because they are camouflaged
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