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3 years ago

PLEASE HELP ME. thx

Mathematics

1 answer:

AfilCa [17]3 years ago

5 0

cant tell direct message me a better picture please

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Pls help its due today if it's correct I'll make you Brainliest!!!!

oee [108]

Answer:

b or c

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please help me out please

mariarad [96]

Answer:

$1.) \\625^2 Ft.\\\\2.)\\306^2Ft.\\\\3.)\\18,750^2 Ft.\\$

Step-by-step explanation:

$1.) \\25*25=625\\\\2.) \\(103*2)+(50*2)=\\206+100=306\\\\3.) \\750*25=18,750$

8 0

3 years ago

Find the sum nineteen and seven tenths added to four and ninety-two hundredths

kirill [66]

The answer is 24.62. Hope this help

8 0

4 years ago

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

IRISSAK [1]

$(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0$

Suppose the ODE has a solution of the form $F(x,y)=C$ , with total differential

$\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

This ODE is exact if the mixed partial derivatives are equal, i.e.

$\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}$

We have

$\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)$

$\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)$

so the ODE is indeed exact.

Integrating both sides of

$\dfrac{\partial F}{\partial x}=(x+y)^2$

with respect to $x$ gives

$F(x,y)=\dfrac{(x+y)^3}3+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2$

$\implies g(y)=-\dfrac{y^3}3-2y+C$

$\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C$

so the general solution to the ODE is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C$

Given that $y(1)=1$ , we find

$\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13$

so that the solution to the IVP is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13$

$\implies\boxed{(x+y)^3-y^3-6y=1}$

5 0

3 years ago

PLSSS HELP MARKING BRAINLIST QUESTION BELOW

Nadya [2.5K]

Answer:

The third answer. ( John's boat was farther from the deck at the beginning but Roberta's boat traveled more quickly)

Step-by-step explanation:

You already know that in the beginning John's boat begins 20 miles from the dock. Now you need to find Roberta! When each boat begins from the dock, it means when x=0. So lets insert that! y= 6(0)+15. You get 15. Now you know that its either the 3rd or 4th answer. The second piece of info you have for John is that it is 45 miles from the dock after 5 hours. Again, in the beginning it started with 20 miles. 45-20= 25 miles TRAVELED in 5 hours. For Roberta, we need to insert 5 in the equation for x. y=6(5)+15. You get: 45 again! But... Roberta started at 15 miles. So 45-15 is 30 miles. She traveled 30 miles in 5 hours. The only answer that is matching to our problem solving is the third answer. Hope this helps!

5 0

4 years ago