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3 years ago

All I have to do is fill in the blanks and I'm a bit confused on what to do. Could someone please help!!

1 answer:

6 0

Jeremy and Randell are brothers and each are trying to raise money for summer camp.
To help Jeremy raise money, his parents told him he could wash each of their cars once a week for $20.00 each. He has already earned $640.00. The football camp that he wants to attend costs $1,469.00.
To help Randell raise money, his parents told him he could mow the grass for them and both sets of grandparents once every 2 weeks and earn $28.00 for each lawn he mows. He has already earned $728.00. The lacrosse camp that he wants to attend costs $1,701.00.
If Jeremy and Randell each earn enough money to attend the camps of their choice, then from this point on (randell) needs to complete (idk what would got here) more chores than(jeremy)

is it asking you to fill in the blanks with their names if so i think it would be this.

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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

Given that

slamgirl [31]

Answer:

P=14

Step-by-step explanation:

Since P depends on two variables x and y as the sum of the two, then;

P = 3+ 11

P = 14

3 0

2 years ago

Find j. 6 + 2(3j – 2) = 4(1 + j)

Sindrei [870]

Answer:

j=1

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Match these similar triangles proof<br>

AlekseyPX

Answer:

Step-by-step explanation:

Ok. They are trying to reconstruct the smaller looking triangle in the bigger triangle using angle A as the common angle.

The first statement is always the given.

Second they constructed line segment XY into the bigger triangle so that XY is parallel to BC.

Third, from the construction of the parallel lines we can now find corresponding angles that are congruent. This would be the use of F.

Since we have all three angles in triangle AXY and triangle ABC, then the construction of the smaller triangle we made inside the bigger triangle is similar to the bigger triangle. So we have the triangles are similar. You could say E or D here in my opinion. This is choice D.

Fifth the creation of those fractions of sides being equal comes from us knowing the corresponding sides of similar triangles are proportional. This is choice H.

Things looked cut off for the sixth thing so I can't fully read it, but it is possible a substitution has occured.

The seventh thing is a congruence statement which can be proven by a congruence postulate. The only one listed is SAS. So that is G.

The last thing, since the triangle construction is congruent to the smaller triangle then we know the smaller triangle is also similar to the bigger triangle since the bigger one is also similar to the construction we made. I really think E and D is interchangeable. Choice E goes here.

7 0

3 years ago

What is one way to test whether an unknown solution is acidic or basic?

jasenka [17]

Answer:

Bases change red litmus into blue and bases have a slippery and soapy texture. Therefore, we can conclude that out of the given options, one way to test whether an unknown solution is acidic or basic is to check whether the solution has a slippery feel

Step-by-step explanation:

3 0

2 years ago