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2 years ago

Please help me I don’t understand this

Mathematics

1 answer:

miss Akunina [59]2 years ago

5 0

Answer:

ok so it's telling you that for 10 folder it cost 1.50 cents each and for 20 it's 3.50 and 30 5$ and 40 6$ and it's basically telling you that what ever your reading you have to think how much did that person spend or bought and how much do u think each folder it's so u have to figure out how many does each cost of cents for each folder.

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The height of a wooden pole, h, is equal to 20 feet. A taut wire is stretched from a point on the ground to the top of the pole.

kykrilka [37]

ANSWER

D. 25 feet

EXPLANATION

The height of the wall,h, the taut wire and the distance from the base of the pole to the point on the ground, formed a right triangle.

According to the Pythagoras Theorem, the sum of the length of the squares of the two shorter legs equals the square of the hypotenuse.

Let the hypotenuse ( the length of the ) taught wire be,l.

Then

${l}^{2} = {h}^{2} + {b}^{2}$

${l}^{2} = {20}^{2} + {15}^{2}$

${l}^{2} = 400 + 225$

${l}^{2} = 625$

$l= \sqrt{625} = 25ft$

8 0

3 years ago

Sigmund wrote four checks last month, and these were the only transactions for his checking account. According to his check regi

Lady bird [3.3K]

Answer:

The Check that hasn't been cleared yet is for $372.15.

Step-by-step explanation:

168.48+ 168.93+ 177.78+ 177.93= 693.12

887.79- 693.12= 194.67

1,065.27- 887.79= 177.48

194.67+ 177.48= 372.15

To check your answer:

372.15+ 693.12= 1,065.27

Hope this helped you!! (:

7 0

3 years ago

Read 2 more answers

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,

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

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