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3 years ago

6

Sianna bought 3 pens, 4 notebooks, and a calculator. The calculator cost $25. If x represents the cost of each pen and y represe

nts the cost of each notebook, which expression represents the amount of change Sianna should receive from $50?
*3x + 4y +25
*50 – (3x + 4y)
*50 – (3x + 4y + 25)
*3x + 4y + 25 + 50

2 answers:

kvv77 [185]3 years ago

8 0

The correct answer would be: 50-(3x+4y+25) because the sum of what's in parethese is what it cost Sianna to buy all of it.

torisob [31]3 years ago

5 0

50-(3x+4y+25) is the answer

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in a certain store, the regular price of a refrigerator is $600. how much money is saved by buying the refrigerator at 20% off t

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60$. These are my equations :
600 * 0.2 = 120 ; 600 - 120 = 480
Then...
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Which then became....
540 - 480 = 60

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2 years ago

Find the missing value.

Fudgin [204]

Answer:

$3125

Step-by-step explanation:

$A = P(1 + \frac{r}{100} )^{n}$

A = 250(1 + 0.25)ⁿ

A = (312.5)ⁿ

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2 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

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3 years ago

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kolezko [41]

Answer: lalalallalalalalal

Step-by-step explanation:

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3 years ago

In parallelogram EFGH, the measure of angle G is (4x + 16)° and the measure of angle H is (2x − 16)°. What is the measure of ang

asambeis [7]

Answer:

m∠H = 44°

Step-by-step explanation:

Consecutive angles of a parallelogram are supplementary. So,

∠H and ∠G are supplementary

4x + 16 + 2x - 16 = 180

6x = 180

x = 30

m∠H = 2x - 16 = 2(30) - 16 = 60 - 16 = 44°

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2 years ago