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Kisachek [45]
3 years ago
6

How many hundredths are in one tenth? Explain using pennies and dimes.

Mathematics
1 answer:
Umnica [9.8K]3 years ago
6 0

Answer:

11 ways

100 Pennies & 0 Dimes

90 Pennies & 1 Dimes

80 Pennies & 2 Dimes

70 Pennies & 3 Dimes

60 Pennies & 4 Dimes

50 Pennies & 5 Dimes

40 Pennies & 6 Dimes

30 Pennies & 7 Dimes

Step-by-step explanation:

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MakcuM [25]
The weight of each cabbage is not given.
I'll assume that each cabbage weighs p pounds.

Now, we know that Yasmin bought 6 cabbage each weighing p pounds. To know the total weight, we will simply multiply the number of cabbages by the weight of each as follows:
Total weight = 6 * p = 6p pounds

Hope this helps :)
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How do you solve equations that contain multiplication or division
alexira [117]
Left to right. Whatever comes first (multiplication or division) you do. This is all part of the PEMDAS/Order of operations.

Hopefully I solved your problem! :)
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3 years ago
HELP ME PLEASE IM ALMOST THERE I LOVE U GUYS I JUST NEED A 100 !!!!!11
Nady [450]
I agree with there answer ^
5 0
3 years ago
If mZDEF = 115, then what are mZFEG and mZHEG? The diagram is not to scale.
stepan [7]

Answer:

the answer is c

Step-by-step explanation:

Im guessing but i hope you get it right im not that smart sooooo ye srry

6 0
3 years ago
38. Evaluate f (3x +4y)dx + (2x --3y)dy where C, a circle of radius two with center at the origin of the xy
lina2011 [118]

It looks like the integral is

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy

where <em>C</em> is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. Then

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on <em>C</em> nor in the region bounded by <em>C</em>, so

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy

where <em>D</em> is the interior of <em>C</em>, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: -2\times \pi\times2^2 = -8\pi.

3 0
3 years ago
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