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Kruka [31]
3 years ago
11

in a plane, line k is parallel to line l and line j is perpendicular to line l. what can you conclude about the relationship bet

ween lines j and k
Mathematics
1 answer:
Sliva [168]3 years ago
4 0
Given that line K is parallel to line L and line J is perpendicular to line L, the relationship between like J and K is also perpendicularity. That is, if lines K and J also meets or intersects at a right angle (90 degrees), similar to how line J and L intersects, then they're said to be perpendicular. 
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​A cable company has a one time installation fee and then a monthly charge to use their service. The total cost of modeled by th
SpyIntel [72]

Answer:

130=monthly charge

119=one time installation fee

Step-by-step explanation:

119 is the y-intercept and in problems like these, it is always the one time fee. 130 is the monthly payment because it has an x, which could account for the number of months. For example , say six months passed, you could multiply 130 * 6 to get the monthly fee for six months. I hope that makes sense!

6 0
2 years ago
Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, i
Doss [256]
C. Hope this helps :)
5 0
3 years ago
Which answer has the smallest value?
Mademuasel [1]

Answer:

The smallest value is 2) the opposite of 50 because it is farthest left of the zero (this makes it the smallest)

Step-by-step explanation:

1) = 100

2) = -50

3) = 0

4) = -25

5) = -5

6 0
3 years ago
Read 2 more answers
Someone please help me
allsm [11]

Answer:

Solutions: x = \frac{-3}{ 4} + i \sqrt{39},  x = \frac{-3}{4} - i \sqrt{39}

Step-by-step explanation:

Given the quadratic equation, 2x² + 3x + 6 = 0, where a =2, b = 3, and c = 6:

Use the <u>quadratic equation</u> and substitute the values for a, b, and c to solve for the solutions:

x = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}

x = \frac{-3 +/- \sqrt{3^{2} - 4(2)(6)} }{2(2)}

x = \frac{-3 +/- \sqrt{9- 48} }{4}

x = \frac{-3 +/- \sqrt{-39} }{4}

x = \frac{-3 + i \sqrt{39} }{4}, x = \frac{-3 - i \sqrt{39} }{4}

Therefore, the solutions to the given quadratic equation are:

x = -\frac{3}{ 4} + i \sqrt{39} ,   x = -\frac{3}{4} - i \sqrt{39}

4 0
2 years ago
Quick help.<br> Will Mark as Brainliest if my laptop stops acting up.
Dimas [21]
Vertex is now at (-1,5)

for
y=a(x-h)^2+k
vertex is (h,k)
so veertex is (-1,5)

y=a(x-(-1))^2+5
y=a(x+1)^2+5
a is a constant, we will asssume that it is 1 because all the choices have 1

y=1(x+1)^2+5
y=(x+1)^2+5

2nd option
8 0
3 years ago
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