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4 years ago

Please help and explain if you can

Mathematics

2 answers:

Keith_Richards [23]4 years ago

7 0

Answer:

I think the answer is 100, but correct me if im wrong

svp [43]4 years ago

3 0

X would be 40 because the angle is inscribed meaning that it’s half the arc since we half 100 and we have half a circle.That means we need the other arc BC to be 80.
Z would be 50 since it’s half the arc which is 100.

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I need help!!!!!!!!!!!!!!!!

Nostrana [21]

Split the figure into 3
area of rectangle is length x width
3x6=18
5x3=15
3x6=18
add them together
18+18+15=51

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3 years ago

Read 2 more answers

a square and rectangle are shown below the width of the rectangle is the same length at a side of the square both represented by

Juli2301 [7.4K]

Answer:

The answer for the length of the rectangle in terms of x is 2x+1=26

Step-by-step explanation:

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3 years ago

PLEASE ANSWER ASAP TIMED! A candle shop owner collected data on customers favorite scents as shown in the table. A customer is r

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Answer: .14

Step-by-step explanation: 2020 edg

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3 years ago

Using the definitions of odd and even functions, explain why y= sin x+1 is neither odd nor even.

Artyom0805 [142]

Answer:

Step-by-step explanation:

The graph is symmetric with respect to the origin therefore it is on odd function. The graph is symmetric to the y- axis therefore it is an even function. The majority of functions are neither odd nor even, however, sine and tangent are odd functions and cosine is an even function.

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4 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago