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SVETLANKA909090 [29]
2 years ago
13

Graph the line based on the following information. y−intercept: 2 Slope: − 4/1 ​

Mathematics
2 answers:
Lena [83]2 years ago
5 0
You would start at 2 and go down 4 and left 1
Neko [114]2 years ago
5 0
As the Y axis, mark a point on the second tick, then go four down and one right and mark your points! Keep doing that and you should have a straight line!
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Evaluate the expression when x = 2, y = 5, and z = 4.
REY [17]

Answer:

9 is the correct answer

(2×2+1)=5÷5×9=9

8 0
2 years ago
Read 2 more answers
Y = 4x - 6
Studentka2010 [4]

Answer:

B) 2x - 12x - 18 = 8

Step-by-step explanation:

2x - 3y = 8

If y = 4x - 6, then substitute and solve:

2x - 3(4x - 6) = 8

2x - 12x - 18 = 8

10x - 18 = 8

10x = 8 + 18

10x = 26

x = 26/10

x = 2.6

y = 4(2.6) - 6

y = 10.4 - 6

y = 4.4

8 0
3 years ago
Help me solve 18.4/2.3*3.4+13.812
dexar [7]
=(8)(3.4)+13.812
=27.2 + 13.812
=41.012
the Answer is 41.012
7 0
2 years ago
for what value of k, the line joining 3x-ky+7=0 is perpendicular to the line joining (4 ,3) and ( 5, -3).
Schach [20]

Answer:

  • k = 18

=========

<h2>Given</h2>

<h3>Line 1</h3>
  • 3x - ky + 7 = 0

<h3>Line 2</h3>
  • Passing through the points (4, 3) and (5, - 3)

<h2>To find</h2>

  • The value of k, if the lines are perpendicular

<h2>Solution</h2>

We know the perpendicular lines have opposite reciprocal slopes, that is the product of their slopes is - 1.

Find the slope of line 1 by converting the equation into slope-intercept from standard form:

<u><em>Info:</em></u>

  • <em>standard form is ⇒ ax + by + c = 0, </em>
  • <em>slope - intercept form is ⇒ y = mx + b, where m is the slope</em>

  • 3x - ky + 7 = 0
  • ky = 3x + 7
  • y = (3/k)x + 7/k

Its slope is 3/k.

Find the slope of line 2, using the slope formula:

  • m = (y₂ - y₁)/(x₂ - x₁) = (-3 - 3)/(5 - 4) = - 6/1 = - 6

We have both the slopes now. Find their product:

  • (3/k)*(- 6) = - 1
  • - 18/k = - 1
  • k = 18

So when k is 18, the lines are perpendicular.

4 0
1 year ago
Match the proper noun to the common noun.
egoroff_w [7]
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3 0
2 years ago
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