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2 years ago

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Mathematics

1 answer:

Kryger [21]2 years ago

6 0

Answer:

Your answer would be

$1 \times 3 = 3$

And

$3 \times 5 = 15$

Step-by-step explanation:

Hope this helps. If I am wrong, REPORT ME PLEASE.

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What is the area of the composite figure?

lana [24]

Hello! :)

$\large\boxed{A = 32.5cm^{2}}$

We can use the following area to solve for the area of the trapezoid pictured:

$A = \frac{1}{2}(b_{1} + b_{2})h$

Where:

h = height

b1 = bottom base

b2 = top base

Plug in the given values:

$A = \frac{1}{2}(9 + 4)(5)$

Simplify:

$A = \frac{1}{2}(65)\\\\A = 32.5 cm^{2}$

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Ivanshal [37]

The answer is -10.
20/-2=-10

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Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

Parameterize the top part of the circle $x^2+y^2=1$ by

$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

with $0\le t\le\pi$ , and the line segment by

$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.

This requires

$\dfrac{\partial f}{\partial x}=\sin x\implies f(x,y)=-\cos x+g(y)$

$\dfrac{\partial f}{\partial y}=\cos y=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\sin y+C$

So we have

$f(x,y)=-\cos x+\sin y+C$

which means $\vec F$ is indeed conservative. By the fundamental theorem, we have

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=f(2,-\pi)-f(1,0)=-\cos2-(-\cos1)=\boxed{\cos1-\cos2}$

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Given ABCE is a rectangle, D is the midpoint of CE<br> Prove AD =~ BD

Bas_tet [7]

Since the congruent operator is â‰… and since AD is congruent to BD, I'm going to assume that you want to prove that AD is congruent to BD.
1. DE is equal to CD by definition since D is the midpoint of CE.
2. AE is equal to BC since opposite sides of a rectangle are equal to each other.
3. Angle AEC is equal to Angle BCE since all angles in a rectangle are right angles and all right angles are equal to each other.
4. Triangles ADE and BDC are congruent to each other because we have SAS congruence for both triangles.
5. AD is congruent to BC since they're corresponding sides of congruent triangles.

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2 years ago