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Aloiza [94]
4 years ago
13

There are some nickels dimes and quarters in a large piggy bank for every two nickels there are 3 dimes for every two dimes that

are 5 quarters there are 500 coins in total how many nickels dimes and quarters are in the piggy bank
Mathematics
1 answer:
creativ13 [48]4 years ago
6 0

We are given

piggy bank has nickels , dimes and quarters

Let's assume

number of nickels =n

every two nickels there are 3 dimes

so, number of dimes are

=\frac{3}{2}n

=1.5n

every two dimes that are 5 quarters there

so, number of quarters are

=\frac{5}{2}\times 1.5n

=3.75n

so, total number of coins = number of nickels + number of dimes +number of quarters

total number of coins =n+1.5n+3.75

there are 500 coins

so, we get

n+1.5n+3.75n=500

now, we can solve for n

6.25n=500

divide both sides by 6.25

so, we get

n=80

number of dimes is 1.5n

=1.5\times 80

=120

number of quarters  is 3.75n

=3.75\times 80

=300

so,

Number of nickels =80

Number of dimes =120

Number of quarters =300............Answer

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Step-by-step explanation:

P=30t-640 main street

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30t-640=25t-450

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Lee sells electronics. He earns a 5% commission on each sale he makes.
NISA [10]

Answer:

a) c = (5/100)*d

b) c = 0.05*d

c) The ratio between what Lee recieves and what he sold.

d) d = 2000

Step-by-step explanation:

Since Lee earns 5% in commission for each sale then:

a) c = (5/100)*d

b) c = 0.05*d

c) In this context the constant of proportionality means the ratio of the value Lee sold that he'll earn as a comission.

d) Since he wants to earn $100, then c = 100 and we solve for d

c = 0.05*d

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4 years ago
Which infinite sequence is generated by the formula an = (-2)^n?
PtichkaEL [24]

Answer:

Step-by-step explanation:

-2,4,-8,16,-32...

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4 0
3 years ago
Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a
NARA [144]

x^3y^2+\sin(x\ln y)+e^{xy}=0

Differentiate both sides, treating y as a function of x. Let's take it one term at a time.

Power, product and chain rules:

\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}

=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}

=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}

Product and chain rules:

\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}

=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)

=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)

=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}

Product and chain rules:

\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}

=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)

=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)

=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0

Isolate the derivative, and solve for it:

\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by y to get rid of that fraction in the denominator.

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}

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Answer:

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Step-by-step explanation:

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