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3 years ago

I’m confused on this one

Mathematics

2 answers:

saul85 [17]3 years ago

8 0

∠1, ∠2, and ∠3 add up to one full turn, or 360°. We can write this fact as the equation

∠1 + ∠2 + ∠3 = 360

∠1 is given to us as a right angle - 90° - and the values ∠2 and ∠3 are given in the question, so we can rewrite our equation with that information as

90 + (x - 7) + 2x = 360

Which we can then solve for x:

x - 7 + 2x = 270 (subtract 90 from both sides)

3x - 7 = 270 (combine like terms)

3x = 277 (add 7 to both sides)

x = 277/3 = 92 1/3 (divide both sides by 3)

Since ∠3 = 2x, we can multiply our result by 2 to find that ∠3 = 2 x (92 1/3) = 184 2/3°

hjlf3 years ago

6 0

All central angles of a circle add to 360 deg.

m<1 = 90 since <1 is a right angle.

m<1 + m<2 + m<3 = 360

90 + x - 7 + 2x = 360

3x + 83 = 360

3x = 277

x = 277/3

m<3 = 2x = 277/3 * 2 = 554/3 = 184 2/3

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Solve the equation.<br> -9x + 1 = -x + 17

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Answer:

The value of $x$ :

$x = -2$

Step-by-step explanation:

Find the value of $x$ :

$-9x + 1 = -x + 17$

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$-9x + 1 + x = -x + x + 17$

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-Subtract both sides by $1$ :

$-8x + 1 - 1 = 17 - 1$

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3 years ago

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For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

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3 years ago

In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p

skelet666 [1.2K]

Answer:

$\mathbf{P(X=5) =0.0888}$

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

$\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}$

$\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}$

where;

n = 8 and π = 0.36

For x = 5

The probability $\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}$

$\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =0.0887645}$

$\mathbf{P(X=5) =0.0888}$ to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5) $\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})$

${P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +$ $\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )$

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1 - P( x ≤ 5 )

P ( x ≥ 6) = 1 - 0.9707

P ( x ≥ 6) = 0.0293

4 0

2 years ago