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Jlenok [28]
2 years ago
11

There is same number of pennies, nickels, and dimes in a wallet. Together they make $1.28. How many coins are there in the walle

t?
Mathematics
2 answers:
Anarel [89]2 years ago
7 0
A = amount

.01a + .05a + .1a = 1.28
Combine like terms.
.16a = 1.28
Divide both sides by .16.
a = 8

8 * 3 = 24

There are 24 coins in the wallet.
Fofino [41]2 years ago
4 0
We call the number of each kind of coins: a
We have the total money is a+ 5a+ 10a= 16a
We have an equation 16a= $1.28 or 16a=128
and we have a= 8
And there are 8x3=24 coins
have fun
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Lisa wants to buy a computer, this computer costs $400, its on sale for 30% off, if you buy it before 9AM to get an additional 1
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Step-by-step explanation:

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3 years ago
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Marrrta [24]

Answer:

Step-by-step explanation:

t = tutoring and g = grocery store

t + g = 15......t = 15 - g

15t + 9g = 159

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-15g + 9g = 159 - 225

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4 0
3 years ago
Consider the differential equation <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D6%20y%5E%7
Nutka1998 [239]
As a Bernoulli equation:

x^2\dfrac{\mathrm dy}{\mathrm dx}=6y^2+6xy\iff x^2y^{-2}\dfrac{\mathrm dy}{\mathrm dx}-6xy^{-1}=6

Let z=y^{-1}\implies\dfrac{\mathrm dz}{\mathrm dx}=-y^{-2}\dfrac{\mathrm dy}{\mathrm dx}. The ODE becomes

-x^2\dfrac{\mathrm dz}{\mathrm dx}-6xz=6
x^6\dfrac{\mathrm dz}{\mathrm dx}+6x^5z=-6x^4
\dfrac{\mathrm d}{\mathrm dx}[x^6z]=-6x^4
x^6z=-6\displaystyle\int x^4\,\mathrm dx
x^6z=-\dfrac65x^5+C
z=-\dfrac6{5x}+\dfrac C{x^6}
y^{-1}=-\dfrac6{5x}+\dfrac C{x^6}
y=\dfrac1{\frac C{x^6}-\frac6{5x}}
y=\dfrac{5x^6}{C-6x^5}

With y(3)=6, we get

6=\dfrac{5(3)^6}{C-6(3)^5}\implies C=\dfrac{4131}2

so the solution is

y=\dfrac{5x^6}{\frac{4131}2-6x^5}=\dfrac{10x^6}{4131-12x^5}

which is valid as long as the denominator is not zero, which is the case for all x\neq\sqrt[5]{\dfrac{4131}{12}}.
8 0
3 years ago
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