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3 years ago

6

HELP ASAP ASAP!! NO LINKS PLEASEEEE OR U WILL BE REPORTED, I NEED AN ANSWER WITH EXPLANATION AND WORK SHOWN OR EXPLAINED, WILL G

IVE BRAINIEST!!

1 answer:

kari74 [83]3 years ago

8 0

Answer:

Method 1 of 7: Squares, Rectangles, and Parallelograms. Measure the width and height.

Step-by-step explanation:

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Hello can anybody answer this to me

Helen [10]

What is this subject

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3 years ago

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Two triangular faces of a rectangular pyramid each have an area of 7.8 cm2 and the other two faces each have an area of 16.5 cm2

Elena L [17]

Answer:

Okay! So to start this problem, we need to look at the given information a little closer to make sure we have everything we need! This is a triangular pyramid, meaning that it has a total of 4 sides, which are all congruent triangles, meaning they all have the same surface area! We are also given the total surface area, which is 2304 cm^2.

If we need to find out how big one of the sides are, we should just have to divide the total surface area by 4, because there are 4 sides with the same surface area! 2304/4 = 576 cm^2, so your answer should be A) 576 cm2!

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4 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

Y=mx+<br> finish the equation<br> Its one of the options<br> M<br> B<br> F<br> C

Dominik [7]

Answer:

b

Step-by-step explanation:

y = mx + b

m is the slope and b is the y-intercept

4 0

3 years ago

Solve for g.<br> f=1/3(g+h-k)<br><br> help it’s timed

andreev551 [17]

Answer:

g = 3f - h + k

Step-by-step explanation:

hope this helps! :D

have a miraculous day, and brainliest is immensely appreciated!! <3

8 0

3 years ago