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Scrat [10]
3 years ago
9

1. Which situation could be represented by y = 4x + 7?

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
5 0

Answer:

  1. B, A package costs $7 plus $4 per pound to ship.
  2. B, y = 2x - 2
  3. B, y = -1/3x + 3
  4. C, y = 2x + 1
  5. B, y = 60x + 40
  6. D, 14 weeks (600 = 40 + 40x)
  7. Sam, ($13hr)

<em>i hope this helps, good luck :)</em>

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Find the slope of the line that passes through (2, 12) and (6, 7).
Elodia [21]

Answer:

m=\frac{1}{4}

Step-by-step explanation:

Gladly!

To find the slope, we need to find the change between <em>point a</em> and <em>point b</em>

With the difference between x1 and x2 being 4, and the difference between y1 and y2 being 1, we get a fraction that looks like this,

\frac{1}{4}

This right here is the slope of the line. Hope this helps!

8 0
3 years ago
2(4x+3)=10x-2<br> Anyone know how to work this out? <br> Working out needed
Step2247 [10]

Answer:

x = 4

Step-by-step explanation:

1. expand.

8x + 6 = 10x - 2

2. Subtract 8x from both sides.

6 = 10x - 2 - 8x

3. simplify 10x - 2 - 8x to 2x - 2.

6 = 2x - 2

4. Add 2 to both sides.

6 + 2 = 2x

5. Simplify 6+2 to 8.

8 = 2x

6. Divide both sides by 2.

\frac{8}{2}  = x

7. Simplify 8/2 to 4.

4 = x

8. switch sides.

x = 4

Therefor, rhe answer is x=4

5 0
3 years ago
Evaluate: ab when a = 2 and b=4
worty [1.4K]

Answer:

8

Step-by-step explanation:

ab = 2 * 4 = 8

7 0
3 years ago
Read 2 more answers
Jk kl and lj are all tangent to o ja = 14 al = 15 and ck=13 find the perimerter of jkl
alexgriva [62]
 JK,KL and LJ are all to O.JA =14, AL=15, and CK=13. Find the perimeter of triangle JKL.? The answer is D. 84

6 0
3 years ago
Read 2 more answers
HELPPP!!!
spayn [35]

Answer:

  A. see below for a graph

  B. f(x, y) = f(0, 15) = 90 is the maximum point

Step-by-step explanation:

A. See below for a graph. The vertices are those defined by the second inequality, since it is completely enclosed by the first inequality: (0, 0), (0, 15), (10, 0)

__

B. For f(x, y) = 4x +6y, we have ...

  f(0, 0) = 0

  f(0, 15) = 6·15 = 90 . . . . . the maximum point

  f(10, 0) = 4·10 = 40

_____

<em>Comment on evaluating the objective function</em>

I find it convenient to draw the line f(x, y) = 0 on the graph and then visually choose the vertex point that will put that line as far as possible from the origin. Here, the objective function is less steep than the feasible region boundary, so vertices toward the top of the graph will maximize the objective function.

8 0
3 years ago
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