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3 years ago

Shannon used equal numbers of quarters and nickels to buy a new notebook that cost $1.50. How many of each coin did she use?

Mathematics

2 answers:

Aleksandr-060686 [28]3 years ago

8 0

U could use a chart and do

N-1

Q-1

T-30

if u kept going 4 later u reach 1.50

ss7ja [257]3 years ago

3 0

Shannon used 5 quarters and 5 nickels.
I basically used the guess-and-check method, and you can verify it by multiplying each coin value by 5.

0.25 x 5 = 1.25
0.05 x 5 = 0.25
1.25 + 0.25 = 1.50

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Will someone help me with both of these I have to solve for the unknown portions please

IceJOKER [234]

Answer:

b=2.5

p=6

Step-by-step explanation:

Take the fraction and simplify it. So 10/8 would be 5/4

You multiply the 2 by 2, giving you h/4

So now you have 5/4=h/4

h=5 but now you have to divide both by 2 again so it can go back to the original fraction.

5÷2=2.5 and 4÷2=2

Second Problem! 4/2 is equal to 2/1

You can also multiply 4 by 3 to get the numerator as 12. Multiply the 4 by 3 and the 2 by 3

Now you have 12/6= 12/p

So p is 6

3 0

3 years ago

Read 2 more answers

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.