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3 years ago

Which graphic organizer is most helpful when comparing and contrasting information?

Mathematics

2 answers:

UNO [17]3 years ago

7 0

Answer:

The answer is the Venn diagram.

Step-by-step explanation:

The graphic organizer that is most helpful when comparing and contrasting information is the Venn diagram.

The Venn diagram can compare and contrast with the help of circles drawn to represent different sets.

The middle overlapping area of circles is used for denoting similarities between sets.

Musya8 [376]3 years ago

5 0

A. a Venn diagram. A Venn diagram is two circles that overlap in the middle. Each circle represents a different subject. The overlapping part is used for comparing, while the rest of the circles are used for contrasting.

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andriy [413]

That is definitely true.

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3 years ago

Captain Tiffany has a ship, the H.M.S. Khan. The ship is two furlongs from the dread pirate Omar and his merciless band of thiev

gavmur [86]

Answer:

1/7

Step-by-step explanation:

The probability that the Captain hits is \dfrac{1}{2}

start fraction, 1, divided by, 2, end fraction.

The probability that the pirate hits is \dfrac{2}{7}

start fraction, 2, divided by, 7, end fraction.

1/2 x 2/7 = 1/7

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3 years ago

Find the value of p in the figure below.

nirvana33 [79]

From the chord theorem we have:

$4\cdot4.5=6\cdot(2p-3)\\\\18=12p-18\\\\12p=18+18\\\\12p=36\qquad|:12\\\\\boxed{p=3}$

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2 years ago

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$

4 0

2 years ago

Change 00345 to a percent.

Wewaii [24]

It would be .345% because to change it to a percent all you have to do is move the decimal point two places to the right.

4 0

3 years ago

Read 2 more answers