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NARA [144]
2 years ago
10

Question 4 of 10

Mathematics
1 answer:
Travka [436]2 years ago
8 0

The variable x represents (d) The width of the deck

<h3>How to determine what x represents?</h3>

The statement is given as:

<em>length of a deck is twice its width</em>

<em />

This means that:

Length = 2 * Width

The equation is given as:

y = 2x

By comparing both equations, we have:

Length = y

Width = x

This means that x represents (d) The width of the deck

Read more about equations at:

brainly.com/question/2972832

#SPJ1

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Use the table to find the relationship between the two quantities.
olchik [2.2K]

Answer:

the answer is 3

Step-by-step explanation:

3 0
3 years ago
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….ummm helppppp please
wel

Answer:

8c proved by me

Step-by-step explanation:

answer is 8c proved

mark me as brainlist

6 0
3 years ago
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If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt
MatroZZZ [7]
Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t
\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle

with 0\le t\le1. Then we find the differential:

\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle
\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt

with 0\le t\le1.

Here, the line integral is

\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle
=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt
=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt
=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt
=x_1y_2-x_2y_1

as required.
4 0
4 years ago
Can you please help me with number 4. 15 points and if you don't know the answer don't say anything please.
iVinArrow [24]
You know the product will be greater than the two factors because multiplication always makes the numbers greater. So 2x12 is 24 and that is greater than the two factors
6 0
3 years ago
Read 2 more answers
Verify sin4x - sin2x = cos4x-cos2x
pychu [463]

Answer:

sin⁴x - sin²x = cos⁴x - cos²x

Solve the right hand side of the equation

That's

sin⁴x - sin²x

<u>From trigonometric identities</u>

<h3>sin²x = 1 - cos²x</h3>

So we have

sin⁴x - ( 1 - cos²x)

sin⁴x - 1 + cos²x

sin⁴x = ( sin²x)(sin²x)

That is

( sin²x)(sin²x)

So we have

( 1 - cos²x)(1 - cos²x) - 1 + cos²x

<u>Expand</u>

1 - cos²x - cos²x + cos⁴x - 1 + cos²x

1 - 2cos²x + cos⁴x - 1 + cos²x

<u>Group like terms</u>

That's

cos⁴x - 2cos²x + cos²x + 1 - 1

<u>Simplify</u>

We have the final answer as

<h3>cos⁴x - cos²x</h3>

So we have

<h3>cos⁴x - cos²x = cos⁴x - cos²x</h3>

Since the right hand side is equal to the left hand side the identity is true

Hope this helps you

4 0
3 years ago
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