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1 year ago

Mr.Simpson feeds his dog 1/2 of a cup of dog food each day. How long will 4 cups of dog food last?

Mathematics

1 answer:

sleet_krkn [62]1 year ago

4 0

Answer:

8 days :)

Step-by-step explanation:

1/2 x 4

1/2 + 1/2 = 1

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8

1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 = 8

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F(x) = x2<br> What is f(x) + f(x) + f(x)?

Yuliya22 [10]

Answer:

if you meant x squared the answer is 3x^2

Step-by-step explanation:

If f(x) is equal to 2x you can substitute x squared for f(x) and that would be x squared+x squared+x squared which is 3x^2

6 0

2 years ago

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Can someone plsss help?

Anni [7]

Answer:

D. $X/Y$

Step-by-step explanation:

4 0

2 years ago

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Can someone pls help me on 4 and 5

never [62]

Answer (4):

a=ub/k because

u=ak/b (starting equation)

ub=ak (multiply b to both sides)

a=ub/k (divide both sides by k)

Answer (6)

x=g+c-y because

g=x-c+y (starting equation)

x+y=g+c (add c to both sides)

x=g+c-y (subtract y from both sides)

5 0

3 years ago

Estimate the sum of the decimal below by rounding to the nearest whole number.

Marizza181 [45]

The non-rounded sum is 15.579, so you’d round up to 16 since 5 and up rounds up. Hope this helps<3

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2 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

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3 years ago