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Lapatulllka [165]
3 years ago
9

How to graph X = (-1/6) y

Mathematics
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer:

X=(-1/6)y

Step-by-step explanation:

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A cone has a height of 1 meter and a diameter of 2 meters. What is it’s volume
Ber [7]

Cone volume formula: V = πr²h/3

r = radius

h = height

The radius is half the diameter, so, we can divide.

2 / 2 = 1

Now, solve with the given values.

V = π(1)²(1)/3

V = π(1)(1/3)

V = 3.14(1/3)

V ≈ 1.05

Therefore, the volume is roughly 1.05m^3

Best of Luck!

6 0
3 years ago
Evaluate the function. f(x) = 3x² – x Find f(10)​
kirill [66]

Answer:

<em>f(10)=290</em>

Step-by-step explanation:

<u>Function Evaluation</u>

Given a function f(x) as an explicit expression, finding f(x=a) implies substituting the value of x by a in every instance it occurs.

We are given the function:

f(x)=3x^2-x

It's required to find f(10). We replace the x for 10:

f(10)=3*10^2-10

Operating:

f(10)=3*100-10

f(10)=300-10

f(10)=290

Thus, f(10)=290

4 0
3 years ago
Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0
natita [175]

Answer:

\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C

Step-by-step explanation:

Given differential equation,

(x^2 + y^2) dx + (x^2 - xy) dy = 0

\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)

Let y = vx

Differentiating with respect to x,

\frac{dy}{dx}=v+x\frac{dv}{dx}

From equation (1),

v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}

v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}

v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}

v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}

x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v

x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}

x\frac{dv}{dx}=\frac{v+1}{v-1}

\frac{v-1}{v+1}dv=\frac{1}{x}dx

Integrating both sides,

\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx

\int{\frac{v-1+1-1}{v+1}}dv=lnx + C

\int{1-\frac{2}{v+1}}dv=lnx + C

v-2ln(v+1)=lnx+C

Now, y = vx ⇒ v = y/x

\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C

5 0
3 years ago
????⃗ (x,y)=(3x−4y)????⃗ +2x????⃗ F→(x,y)=(3x−4y)i→+2xj→ and ????C is the counter-clockwise oriented sector of a circle centered
Karolina [17]

By Green's theorem, the integral of \vec F along C is

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial(2x)}{\partial x}-\frac{\partial(3x-4y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=6\iint_D\mathrm dx\,\mathrm dy

which is 6 times the area of D, the region with C as its boundary.

We can compute the integral by converting to polar coordinates, or simply recalling the formula for a circular sector from geometry: Given a sector with central angle \theta and radius r, the area A of the sector is proportional to the circle's overall area according to

\dfrac A{\frac\pi3\,\rm rad}=\dfrac{16\pi}{2\pi\,\rm rad}\implies A=\dfrac{8\pi}3

so that the value of the integral is

\dfrac{6\times8\pi}3=\boxed{16\pi}

6 0
3 years ago
Which of the following statements is not true?
Nat2105 [25]
X5y5z5 = xyz5 is not true
6 0
3 years ago
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