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slamgirl [31]
2 years ago
13

In 2000, Carl’s Candy Shop sold gummy bears for $0. 38 per pound, and in 2008 they cost $0. 72 per pound. Let y represent the co

st per pound of the gummy bears and x represent the years after 2000
Mathematics
1 answer:
tamaranim1 [39]2 years ago
3 0

The linear function that represent the cost per pound of the gummy bears in x years after 2000 is given by:

y(x) = 0.0425x + 0.38

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = mx + b

In which:

  • m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
  • b is the y-intercept, which is the value of y when x = 0, which can also be interpreted as the initial value of the function.

In this problem, we have that:

  • In 2000, Carl’s Candy Shop sold gummy bears for $0. 38 per pound, hence b = 0.38.
  • In 2008 they cost $0.72 per pound, that is, in 8 years there was an increase of $0.34 per pound, hence the slope is m = 0.34/8 = 0.0425.

Thus, the function is:

y(x) = 0.0425x + 0.38

More can be learned about linear functions at brainly.com/question/24808124

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Adding fractions<br> 5 over 10 + 1 over 3
Ray Of Light [21]

Answer:

25/30

Simplified: 5/6

Step-by-step explanation:

Have a good day!

Please rate and mark brainliest!

7 0
2 years ago
Read 2 more answers
Please help me will mark the brainliest
coldgirl [10]

Answer:

The answer is A)9.89cm squared

Step-by-step explanation:

Split the shape into a triangle and a square. The square's area is 2.6 × 2.3 which equals 5.98. The triangle's area is 6-2.6=3.4, 3.4×2.3= 7.82. But since its a triangle, divide it by two. 7.82÷2=3.91. Now add both of the found areas to eachother, 3.91+5.98=9.98 cm squared. Hope that makes sense and helps!

7 0
3 years ago
Read 2 more answers
What is the functions average rate of change over each interval?
telo118 [61]

we can use average rate of change formula

\frac{f(b)-f(a)}{b-a}

(a) x = -3 to x = -2

Firstly, we will find points

(-3, 0) and (-2,2)

now, we can plug these values into formula

and we get

=\frac{2-0}{-2+3}

=2

(b)x = -2 to x =1

Firstly, we will find points

(-2,2) and (1,-4)

now, we can plug these values into formula

and we get

=\frac{-4-2}{1+2}

=-2

(c) x = 0 to x =1

Firstly, we will find points

(0,-3) and (1,-4)

now, we can plug these values into formula

and we get

=\frac{-4+3}{1-0}

=-1

(d) x = 1 to x =2

Firstly, we will find points

(1,-4) and (2,0)

now, we can plug these values into formula

and we get

=\frac{0+4}{2-1}

=4

(e) x = -1 to x =0

Firstly, we will find points

(-1,0)  and (0,-3)

now, we can plug these values into formula

and we get

=\frac{-3-0}{0+1}

=-3


(f) x = -1 to x =2

Firstly, we will find points

(-1,0)  and (2,0)

now, we can plug these values into formula

and we get

=\frac{0-0}{2+1}

=0


3 0
3 years ago
The ordered pair(5,-3) is a solution to which of the following inequalities? a. y≥−2x+8 b. −2y&lt;3x−9 c. y−2x&gt;5 d. 4y+2x≤−1
stiks02 [169]

Answer:

Option d. 4y+2x\leq -1

Step-by-step explanation:

we know that

If a ordered pair is a solution of an inequality, then the ordered pair must be satisfy the inequality

Verify each case

case a) y\geq2x+8

we have

x=5, y=-3

Substitute the value of x and the value of y in the inequality and then compare the results

-3\geq2(5)+8

-3\geq18 -----> is not true

therefore

the ordered pair is not a solution

case b) -2y

we have

x=5, y=-3

Substitute the value of x and the value of y in the inequality and then compare the results

-2(-3)

6 -----> is not true

therefore

the ordered pair is not a solution

case c) y-2x>5

we have

x=5, y=-3

Substitute the value of x and the value of y in the inequality and then compare the results

-3-2(5)>5

-13>5 -----> is not true

therefore

the ordered pair is not a solution

case d) 4y+2x\leq -1

we have

x=5, y=-3

Substitute the value of x and the value of y in the inequality and then compare the results

4(-3)+2(5)\leq -1

-2\leq -1 -----> is  true

therefore

the ordered pair is a solution

7 0
4 years ago
Can something help me with a and b
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A. 399
b. charlotte ....
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