A music store receives shipments of recorders and harmonics. A shipment of 5

2 answers:

8 0

Harmonica = $8
Recorder = $6

We’ll call r the cost of a recorder, h the cost of a harmonica.

5r + 4h = 62
10r + 3h = 84

First, we’ll solve for r in terms of h:
10r = 84 -3h
r = 8.4 - 0.3h

Then we’ll substitute that expression for r in the other equation:

5(8.4 - 0.3h) + 4h = 62
42 - 1.5h + 4h = 62
2.5h = 20
h = 8

Then we’ll substitute that in to solve for the numerical value of r

r = 8.4 - 0.3h
r = 8.4 - 0.3(8)
r = 8.4 - 2.4
r = 6

Check the math:
5r + 4h = 62
5(6) + 4(8) = 30 + 32 = 62

Let me know if you have any questions.

Varvara68 [4.7K]2 years ago

4 0

Answer:

Recorders= $6

Harmonics= $8

Step-by-step explanation:

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Somebody who knows how to do this?? plz answer all the questions correctly thanks!

Ksju [112]

Answer:

a) 5/21

b) 4/21

c) 4/21

d) 8/21

Step-by-step explanation:

total number of coins: 21

a) number if dollars: 5

therefore fraction is 5/21

b) number of quarters: 4

therefore fraction is 4/21

c) number of dimes: 4

therefore fraction is 4/21

d) number of nickels: 8

therefore fraction is 8/21

4 0

3 years ago

If g (x) = 1/x then [g (x+h) - g (x)] /h

lys-0071 [83]

Answer:

$\dfrac{-1}{x(x+h)}, h\ne 0$

Step-by-step explanation:

If $g(x) = \dfrac{1}{x}$ , then $g(x+h) = \dfrac{1}{x+h}$ . It follows that

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}$

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}$