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3 years ago

What is the correct definition of acceleration?

Mathematics

1 answer:

EastWind [94]3 years ago

5 0

The answer would be C. The change of Velocity over time. :)

You might be interested in

Find twenty rational number between -3/7 and 2/3

horrorfan [7]

Answer:

-2/7,-1/7,-2/5, -1/3, -1/10,-1/9,-1/8,-1/11,-1/12,0,1/100, 2/99,1/3,1/4,1/5,3/7,3/11,4/21, 5/23, 11/109,1/2

Step-by-step explanation:

all the number that can be written as a fractions betwwen -3/7 and 2/3

-3/7≤ x/y≤ 2/3

-3/7 ≈ -0.428..

2/3 ≈ 0.667

8 0

2 years ago

Solve the proportion x+1/x-1 = 14/20

polet [3.4K]

Cross multiply
20(x+1)=14(x-1)
20x+20=14x-14
Subtract 14x on each side giving you
6x+20=-14
Subtract 20 on each side giving you
6x=-34
x=-34/6. Simplifying that you get x=-17/3. I hope this helps!!!

4 0

3 years ago

Read 2 more answers

(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

234 students in 9 different classrooms. What is the ratio of students to classrooms?

Vilka [71]

Answer:

26 students per classroom

Step-by-step explanation:

To solve this, the ratio 234 students:9 classrooms must be converted to:

s students:1 classroom

To convert the ratio, 9 was divided by 9 to get 1, and so you must also divide 234 by 9 to maintain the same ratio, which gives us the answer:

26 students: 1 classroom

5 0

2 years ago

Why are astrologers not considered to be scientists?

Pachacha [2.7K]

The answer is: [C]: "Astrologers do not use the scientific method."
_____________________________________________________
Note: Choice [A] is incorrect. Astrologers do use tools.
_____________________________________________________
Choice [B] is incorrect. Some astrologers do observe the universe.
_____________________________________________________
Choice [D] is incorrect. Astrologers may or may not go to college; and whether or not they do or do not is irrelevant as regard to whether to they are considered "scientists".
_____________________________________________________

8 0

2 years ago

Read 2 more answers