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krek1111 [17]
3 years ago
15

Pls help me guys I really need it, pls dont say anything random!!!​

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
4 0

Answer:

sorry but you need to answer that in your own dont be lazy

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How can you plot the reflection of point A across the y-axis? Give the coordinates
Maslowich
Take the points you already plotted and plot them on the other side, regardless if the number is negative or nit
8 0
4 years ago
-5-3x ≥ 2(10+2x)+3<br> Can you solve this as an inequality?
Thepotemich [5.8K]

Answer: x ≤ -4

Step-by-step explanation: The explination  will be placed in the comment area :)

5 0
3 years ago
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Given the force field F, find the work required to move an object on the given oriented curve. F = (y, - x) on the path consisti
timofeeve [1]

Answer:

0

Step-by-step explanation:

We want to compute the curve integral (or line integral)

\bf \int_{C}F

where the force field F is defined by

F(x,y) = (y, -x)

and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9).

We can write  

C = \bf C_1+C_2

where  

\bf C_1 =  line segment from (1, 5) to (0, 0)  

\bf C_2 = line segment from (0, 0) to (0, 9)

so,

\bf \int_{C}F=\int_{C_1}F+\int_{C_2}F

Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with

<em>r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1 </em>

Hence \bf C_1 can be parameterized as

\bf r_1(t) = (1-t, 5-5t) for 0 ≤ t ≤ 1

and \bf C_2 can be parameterized as

\bf r_2(t) = (0, 9t) for 0 ≤ t ≤ 1

The derivatives are

\bf r_1'(t) = (-1, -5)

\bf r_2'(t) = (0, 9)

and

\bf \int_{C_1}F=\int_{0}^{1}F(r_1(t))\circ r_1'(t)dt=\int_{0}^{1}(5-5t,t-1)\circ (-1,-5)dt=0

\bf \int_{C_2}F=\int_{0}^{1}F(r_2(t))\circ r_2'(t)dt=\int_{0}^{1}(9t,0)\circ (0,-9)dt=0

In consequence,

\bf \int_{C}F=0

6 0
4 years ago
Look at the question below.
Crank

Answer:

A is correct answer

Step-by-step explanation:

If you add the two b's then it would be "2b" which is equivalent to the the other problem on the other side of the equal sign which is 2b + 3c

8 0
3 years ago
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Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.
rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed  as:

y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}

Thus, the area of the surface is as follows:

\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA

= \iint_D \sqrt{x^2 +y^2 +1 } \ dA

= \int^{2 \pi}_{0} \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr \ d \theta

=2 \pi \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr

= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0

= 2 \pi \times \dfrac{1}{3}  \Bigg [ (37)^{3/2} - 1 \Bigg]

= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]

3 0
3 years ago
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