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Leya [2.2K]
3 years ago
15

Suppose j varies jointly with g and​ v, and j=2 when g=4 and v=3. Find j when g=8 and v=9.

Mathematics
1 answer:
vladimir2022 [97]3 years ago
3 0

Answer:

j=12

Step-by-step explanation:

we know that

In a join variation, If j varies jointly with respect to g and v, the equation will be of the form j = kgv

where k is a constant

step 1

Find the value of k

we have

j=2,g=4,v=3

substitute and solve for k

2 = k(4)(3)

2 = 12k

k=\frac{1}{6}

The equation is equal to

j=\frac{1}{6}gv

step 2

Find the value of j when g=8,v=9

substitute the values in the equation and solve for j

j=\frac{1}{6}(8)(9)

j=12

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Hope this helps :)

8 0
2 years ago
Rosa earns $12 per hour. In 7.5 hours, she will earn x dollars. Which is a valid proportion that can be used to solve the proble
Tamiku [17]

Answer:

StartFraction x over 7.5 EndFraction = StartFraction 12 over 1 EndFraction

Step-by-step explanation:

Here, in this question, given that Rosa earns $12 in an hour, she earns x dollars in 7.5 hours. So we are interested in finding the value of x.

Let’s write this in terms of proportion;

$12 = 1 hour

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$12 * 7.5 hours = $x * 1 hour

So this can be rewritten as;

$x/7.5 = $12/1 hour

5 0
3 years ago
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Answer:

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3 0
2 years ago
A jeweler wants to minimize business costs and has found that the average cost in dollars per necklace is given by C(x)=0.5x^2-7
Vika [28.1K]

Answer:

The maximum cost is 41 dollars per necklace.

Step-by-step explanation:

The average cost in dollars/necklace is given by :

C(x)=0.5x^2-7x+65.5  .....(1)

Where

x is the number of necklaces that are created

To maximize the cost, find the value of x by putting dC/dx = 0

\dfrac{d}{dx}(0.5x^2-7x+65.5)=0\\\\x-7=0\\\\x=7

Now put the value of x = 7 in equation (1).

C(x)=0.5(7)^2-7(7)+65.5\\\\=41\ \text{dollars per necklace}

Hence, the maximum cost is 41 dollars per necklace.

8 0
2 years ago
10 points! Please help ASAP.
DerKrebs [107]
Answer: Yes
Explanation:
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If Maxine is correct, then she spent 1.2
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15 ⋅ 1.2 = 18.0 = 18

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6 0
3 years ago
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