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3 years ago

What is 2,606 + 7,025?

Mathematics

2 answers:

trapecia [35]3 years ago

6 0

Answer: 9631

2606

+7025

9631

cestrela7 [59]3 years ago

5 0

Answer:

2606 + 7025 = 9631

Pls mark me brainliest

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Nadia runs 8 miles each weekend. How many yards does Nadia run each weekend

luda_lava [24]

Answer:

14080 yds

Step-by-step explanation:

We know that 1 mile = 1760 yards

8 miles * 1760 yds/ 1 mile = (8*1760) yds=14080 yds

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3 years ago

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Megan has a piece of fabric which is a right triangular shape. The length of leg 1 is 4 m. The length of the other leg is twice

Bess [88]

Answer:

8.94 m

Step-by-step explanation:

Use the Pythagorean Theorum - $a^2 + b^2 = c^2$ . Use leg 1 as a, leg 2 as b, and the third side as c.

We know leg 1 is 4, and that leg 2 is twice as long as leg 1...

$(4)^2 + (4*2)^2 = c^2$

Now, we use the Order of Operations to find $c^2$ ...

$4^2+8^2=c^2\\16+64=c^2\\80=c^2$

Now, we find the square root of $c^2$ to find c...

$\sqrt{80} =8.94$

So, the length of the third side is 8.94 m.

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3 years ago

George weighs twice as much as his little brother Sam.

german

74 pounds because sam weight 37
37x2 = 74

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3 years ago

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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.