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kherson [118]
2 years ago
15

GOT AN Y INTERCEPT AND GRAPH PROBLEM PLS HELPPPPPP!!! HOPE it's NOT TO HARD TO READDDDDDDDD

Mathematics
1 answer:
Novay_Z [31]2 years ago
4 0

Answer:

1) It has a y-intercept of 4

2) A, the rate of change represents how fast the boys run every hour.

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O Mr. Math S. Fun is looking for his missing pair of
dalvyx [7]

Answer:

1/3

Step-by-step explanation:

add all of the totals including the missing pair

we now have 24

7+1 is 8

24 minus 8 is 16

24 can be split into 2 groups of 8

and we have 2/3 left. which makes our solution 1/3

3 0
2 years ago
13/100 + 11/10 helpppp
zhannawk [14.2K]

Answer:

13/100 + 11/10 = 123/100

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
PLEASE ANSWER ASAP!!!
Vesna [10]

Answer:

C=x (x+3)

Step-by-step explanation:

x cannot divide x+3 definitely so the denominators must be multiplied to get the least common denominator.

4 0
3 years ago
Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find
Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk x^2+y^2\le3, I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere x^2+y^2+z^2=3.

a. Let C denote the hemispherical <u>c</u>ap z=\sqrt{3-x^2-y^2}, parameterized by

\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k

with 0\le u\le2\pi and 0\le v\le\frac\pi2. Take the normal vector to C to be

\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k

Then the upward flux of \vec F=(2z+2)\,\vec k through C is

\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du

\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du

=\boxed{2(3+2\sqrt3)\pi}

b. Let D be the disk that closes off the hemisphere C, parameterized by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath

with 0\le u\le\sqrt3 and 0\le v\le2\pi. Take the normal to D to be

\vec s_v\times\vec s_u=-u\,\vec k

Then the downward flux of \vec F through D is

\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv

=\boxed{-6\pi}

c. The net flux is then \boxed{4\sqrt3\pi}.

d. By the divergence theorem, the flux of \vec F across the closed hemisphere H with boundary C\cup D is equal to the integral of \mathrm{div}\vec F over its interior:

\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV

We have

\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2

so the volume integral is

2\displaystyle\iiint_H\mathrm dV

which is 2 times the volume of the hemisphere H, so that the net flux is \boxed{4\sqrt3\pi}. Just to confirm, we could compute the integral in spherical coordinates:

\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi

4 0
3 years ago
In triangle opq right angled at p op=7cm,oq-pq=1 determine the values of sinq and cosq
soldi70 [24.7K]

Answer:

see explanation

Step-by-step explanation:

let pq = x

given oq - pq = 1 then oq = 1 + x

Using Pythagoras' identity, then

(oq)² = 7² + x²

(1 + x)² = 49 + x² ( expand left side )

1 + 2x + x² = 49 + x² ( subtract 1 from both sides )

2x + x² = 48 + x² ( subtract x² from both sides )

2x = 48 ( divide both sides by 2 )

x = 24 ⇒ pq = 24

and oq = 1 + x = 1 + 24 = 25 ← hypotenuse

sinq = \frac{opposite}{hypotenuse} = \frac{7}{25}

cosq = \frac{adjacent}{hypotenuse} = \frac{24}{25}



6 0
3 years ago
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