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notka56 [123]
3 years ago
5

If DF = 170, find the value of x. Then find DE and EF. DE=3x + 20 EF=2x + 40

Mathematics
1 answer:
Julli [10]3 years ago
7 0

9514 1404 393

Answer:

  DE = 86

  EF = 84

Step-by-step explanation:

We assume that point E lies on segment DF, so that ...

  DE + EF = DF

  (3x +20) +(2x +40) = 170

  5x = 110 . . . . . . . . . . . . . . . collect terms, subtract 60

  x = 22 . . . . . . . . . . . . divide by 5

  DE = 3×22 +20 = 66 +20 = 86

  EF = 2×22 +40 = 44 +40 = 84

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67. The line contains the point (4,0) and is parallel<br> to the line defined by 3x = 2y.
olganol [36]

Answer:

y=\frac{3}{2} x-6

Step-by-step explanation:

Hi there!

<u>What we need to know:</u>

  • Linear equations are typically organized in slope-intercept form:
  • y=mx+b where m is the slope of the line and b is the y-intercept (the value of y when the line crosses the y-axis)
  • Parallel lines will always have the same slope but different y-intercepts.

<u>1) Determine the slope of the parallel line</u>

Organize 3x = 2y into slope-intercept form. Why? So we can easily identify the slope, m.

3x = 2y

Switch the sides

2y=3x

Divide both sides by 2 to isolate y

\frac{2y}{2} = \frac{3}{2} x\\y=\frac{3}{2} x

Now that this equation is in slope-intercept form, we can easily identify that \frac{3}{2} is in the place of m. Therefore, because parallel lines have the same slope, the parallel line we're solving for now will also have the slope \frac{3}{2} . Plug this into y=mx+b:

y=\frac{3}{2} x+b

<u>2) Determine the y-intercept</u>

y=\frac{3}{2} x+b

Plug in the given point, (4,0)

0=\frac{3}{2} (4)+b\\0=6+b

Subtract both sides by 6

0-6=6+b-6\\-6=b

Therefore, -6 is the y-intercept of the line. Plug this into y=\frac{3}{2} x+b as b:

y=\frac{3}{2} x-6

I hope this helps!

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Write the equation of the line that is perpendicular to 3x+2y=8 and goes through the point (-5,2)
kifflom [539]

Answer:

\displaystyle y=\frac{2}{3}(x+5)+2

Step-by-step explanation:

We need to find the equation of the line perpendicular to the line 3x+2y=8 and passes through (-5,2).

The given line can be expressed as:

\displaystyle y=-\frac{3}{2}x+4

We can see the slope of this line is m1=-3/2.

The slopes of two perpendicular lines, say m1 and m2, meet the condition:

m_1.m_2=-1

Solving for m2:

\displaystyle m_2=-\frac{1}{m_1}

\displaystyle m_2=-\frac{1}{-\frac{3}{2}}

\displaystyle m_2=\frac{2}{3}

Now we know the slope of the new line, we use the slope-point form of the line:

y=m(x-h)+k

Where m is the slope and (h,k) is the point. Using the provided point (-5,2):

\boxed{\displaystyle y=\frac{2}{3}(x+5)+2}

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