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3 years ago

Help please!!!!!.....

Mathematics

2 answers:

GarryVolchara [31]3 years ago

6 0

Answer:

294

Step-by-step explanation:

7*7*6=294

svetlana [45]3 years ago

4 0

L•w•h (length times width times height)
7•7•6
=49•6
=294

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What is the equation of the line graphed below?

Ray Of Light [21]

Answer:

The equation is y = 2x. The slope is two and the line is a direct variation

5 0

3 years ago

Consider the function f(x)=−2x3+2x2−2x+2 Find the average slope of this function on the interval (4,9). By the Mean Value Theore

____ [38]

Answer:

Average slope =-242

c=6.67

Step-by-step explanation:

We are given that a function

$f(x)=-2x^3+2 x^2-2 x+2$ and interval (4,9)

We have to find the average slope of the given function and value of c in the given interval.

Using mean value theorem

${f(b)-f(a)}{b-a}=f'(c)$

a=4 and b=9

$f(9)=-2(9)^3+2(9)^2-2(9)+2$

f(9)=-1312

$f(4)=-2(4)^3+2(4)^2-2(4)+2$

$f(4)=-128+32-8+2=-102$

Substitute the values then we get

$f'(c)=\frac{f(9)-f(4)}{9-4}=\frac{-1312+102}{5}=-242$

Hence, the average value of slope is -242.

We know that f'(c)=-242

$f'(x)=-6x^2+4x-2$

Substitute x=c

Then $f'(c)=-6c^2+4c-2$

$-6 c^2+4 c-2=-242$

$-3 c^2+ 2c -1=-121$

Dividing on both sides by 2

$3c^2-2c+1-121=0$

$3c^2-2 c-120=0$

$3c^2-20c+18c-120=0$

$(3c-20)(c+6)=0$

$3c=20$ and $c+6=0$

$c=\frac{20}{3}$ and c=-6 It is not possible because it does not lie in the given interval

Therefore,c=6.67 lie in the given interval

4 0

2 years ago

To take a taxi, it costs \$3.00$3.00dollar sign, 3, point, 00 plus an additional \$2.00$2.00dollar sign, 2, point, 00 per mile t

djverab [1.8K]

Answer 8 miles

Step-by-step explanation:

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3 years ago

(X^2+y^2+x)dx+xydy=0 Solve for general solution

aksik [14]

Check if the equation is exact, which happens for ODEs of the form

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

We have

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

$N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y$

so the ODE is not quite exact, but we can find an integrating factor $\mu(x,y)$ so that

$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$

is exact, which would require

$\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$

$\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}$

Notice that

$\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y$

is independent of x, and dividing this by $N(x,y)=xy$ gives an expression independent of y. If we assume $\mu=\mu(x)$ is a function of x alone, then $\frac{\partial\mu}{\partial y}=0$ , and the partial differential equation above gives

$-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}$

which is separable and we can solve for $\mu$ easily.

$-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}$

$\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x$

$\ln|\mu|=\ln|x|$

$\implies \mu=x$

So, multiply the original ODE by x on both sides:

$(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0$

Now

$\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy$

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

so the modified ODE is exact.

Now we look for a solution of the form $F(x,y)=C$ , with differential

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

The solution F satisfies

$\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2$

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

Integrating both sides of the first equation with respect to x gives

$F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)$

Differentiating both sides with respect to y gives

$\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y$

$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

So the solution to the ODE is

$F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C$

$\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}$

5 0

3 years ago

Please help meee!!!lol

mihalych1998 [28]

Answer:

X represents the number of product so 3 cause you have 2 pastas and 1 cheese. The length represents the total cost. I think 3 would represent cost? I'm not totally sure on that one. Hope this helps!

Step-by-step explanation:

7 0

2 years ago