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Goryan [66]
3 years ago
11

what is the answer to the volleyball team at blue point high school has saved $680 and the team plans to spend no more then that

amount on balls and nets a ball is $6 and a net is $23 write the inequality that represents the combinations of x balls and y nets that can be purchased
Mathematics
1 answer:
zloy xaker [14]3 years ago
4 0

Answer:

6 x + 23 y  < $ 680  represents the combinations of x balls and y nets that can be purchased.

Step-by-step explanation:

The number of balls than can be purchased  = x

The number of nets than can be purchased  = y

Total savings  = $ 680

Now, the cost of 1 ball =$6

So, the cost of x balls = x ( cost of 1 ball) = x ( $6)  = 6 x

Also, the cost of 1 net =$23

So, the cost of y nets = y ( cost of 1 net = y ( $23)  = 23 y

So, the total amount spent on x balls  and y nets

= Amount spent on x balls + Amount on y nets

=  6 x + 23 y

Also, the saved amount is $680

⇒<u> 6 x + 23 y  < $ 680</u>

Hence, 6 x + 23 y  < $ 680  represents the combinations of x balls and y nets that can be purchased.

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The graph of linear function k passes through the points (-3, 0) and (1, 8). Which statement must be true?
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Answer:

  • B. The slope of the graph of k is 2.

Step-by-step explanation:

<h3>Find the equation of the line.</h3>

<u>The slope is:</u>

  • m = (8 - 0)/(1 + 3) = 2

<u>The y-intercept is:</u>

  • 0 = -3*2 + b
  • b = 6

<u>The line is:</u>

  • y = 2x + 6
<h3>Now lets look at the answer options</h3>

A. <u>The x-Intercept of the graph of k is 8.</u>

  • The x- intercept is -3, False

B. <u>The slope of the graph of k is 2.</u>

  • m = 2, True

C. <u>The graph of k passes through (-1, -8).</u>

  • -8 = 2(-1) + 6
  • -8 = 4, False

D. <u>The zero of k is 3.</u>

  • x = 0 ⇒ y = 0+ 6 = 6,  False
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Jim bought 20 rosebushes last week. the regular price is $5.00 per bush, but last week they were on sale at 2 for $8.50. how muc
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Select ALL the correct answers.<br> Select all functions that have a y-intercept of (0,5).
otez555 [7]
1.) f(x)=7(b)^x-2
x=0→f(0)=7(b)^0-2=7(1)-2=7-2→f(0)=5→(x,f(x))=(0,5) Ok

2.) f(x)=-3(b)^x-5
x=0→f(0)=-3(b)^0-5=-3(1)-5=-3-5→f(0)=-8→(x,f(x))=(0,-8) No

3.) f(x)=5(b)^x-1
x=0→f(0)=5(b)^0-1=5(1)-1=5-1→f(0)=4→(x,f(x))=(0,4) No

4.) f(x)=-5(b)^x+10
x=0→f(0)=-5(b)^0+10=-5(1)+10=-5+10→f(0)=5→(x,f(x))=(0,5) Ok

5.) f(x)=2(b)^x+5
x=0→f(0)=2(b)^0+5=2(1)+5=2+5→f(0)=7→(x,f(x))=(0,7) No

Answers:
First option: f(x)=7(b)^x-2
Fourth option: f(x)=-5(b)^x+10 
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