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2 years ago

forty percent of the dogs have been adopted ( there are four dogs) How many of the dogs have been adopted

Mathematics

1 answer:

Mashutka [201]2 years ago

8 0

100%-40%=0.6 (60%) so if 40% were adopted then and only 4 left then

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C/5 - 4= -3 pls help!

olya-2409 [2.1K]

Answer:

The value of c from the equation is 5.

Step-by-step explanation:

c/5 - 4 = -3

Add 4 to both sides of the equation.

c/5 = 1

Multiply both sides by 5 to get c by itself.

c = 5

4 0

2 years ago

Read 2 more answers

Please helppppp pls it's in the screenshot

sergey [27]

Answer:

4. 11/6, 121/36

5. 5/8, 25/64

6. 4/7, 16/49

7. 14/9, 196/81

Step-by-step explanation:

The ratio of perimeters is the same as the ratio of corresponding legs:

perimeter ratio = red leg/blue leg

The area ratio is the square of that:

area ratio = (perimeter ratio)²

perimeter ratio = 11/6

area ratio = (11/6)² = 121/36

perimeter ratio = 5/8

area ratio = (5/8)² = 25/64

perimeter ratio = 4/7

area ratio = (4/7)² = 16/49

perimeter ratio = 14/9

area ratio = (14/9)² = 196/81

3 0

1 year ago

florida received 15 inches of rainfall in 3 months of the year. If the state receive 4.25 each month until the end of the year,

kherson [118]

Since there are 12 months in there year and there have already been 3 months there are 9 left

we need to multiply 4.25 by 9 and then add the first 15 inchess

4.25 times 9 = 38.25+15 = 53.25

Florida will recieve 53.25 inches of rain

I hope I've helped!

3 0

2 years ago

Aidan bought a jacket on sale for $38.40. If the original price of the jacket was $64, what was the percent of the discount?

UkoKoshka [18]

Answer:

60%

Step-by-step explanation:

$64*.60= $38.40

3 0

2 years ago

Read 2 more answers

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

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

2 years ago