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2 years ago

What Two pairs of fractions are shown. Which symbol correctly compares 2/3 2/4 each pair of fractions?

Mathematics

1 answer:

VikaD [51]2 years ago

4 0

Answer:

Greater than symbol.

2/3 > 2/4

Step-by-step explanation:

We have two fractions, 2/3 and 2/4

We then go back to the inequality part of mathematics where we have symbols such as greater than (>), less than (<), and equal to (=).

Of the three inequality symbols, we look and pick at the one that represents best, the fractions we were given.

2/3 and 2/4

With 2/3 being greater than 2/4.

2/3 = 0.67

2/4 = 0.50

Succinctly, 2/3 > 2/4.

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Write each expression as a single power of 10

swat32

Answer:

It is

${10}^{5}$

Step-by-step explanation:

$\frac{ {10}^{4}. {10}^{5} . {10}^{6} }{ {10}^{3} . {10}^{7} } = \frac{ {10}^{(4 + 5 + 6)} }{ {10}^{(3 + 7)} } \\ \\ = \frac{ {10}^{15} }{ {10}^{10} } \\ \\ = {10}^{(15 - 10)} \\ = {10}^{5}$

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3 years ago

What is the area of the irregular figure below?

expeople1 [14]

Answer:

360 centimeters squared

Step-by-step explanation:

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2 years ago

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Rectangle ABCD was dilated to create rectangle A'B'C'D. Triangle A B C D is dilated to form triangle A prime B prime C prime D.

Julli [10]

Answer:

$AB = 6$

Step-by-step explanation:

Given

$BC = 3.8$

$A'B' = 15$

$B'C' = 9.5$

Required

Determine the length of AB

Because A'B'C'D is a dilation of ABCD, then the following relationship must exist:

$A'B' : AB = B'C' : BC$

Substitute values for A'B', B'C' and BC

$15 : AB = 9.5 : 3.8$

Express the ratio as fractions:

$\frac{AB}{15} = \frac{3.8}{9.5}$

Multiply through by 14

$15 * \frac{AB}{15} = \frac{3.8}{9.5} * 15$

$AB = \frac{3.8}{9.5} * 15$

$AB = \frac{3.8* 15}{9.5}$

$AB = \frac{57}{9.5}$

$AB = 6$

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2 years ago

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Find all solutions of the equation in the interval [0, 2pi).

natali 33 [55]

Answer:

Step-by-step explanation:

Begin by squaring both sides to get rid of the radical. Doing that gives you:

$sin^2x=1-cosx$

Now use the Pythagorean identity that says

$sin^2x =1-cos^2x$ and make the replacement:

$1-cos^2x=1-cosx$ . Now move everything over to one side of the equals sign and set it equal to 0 so you can factor:

$1-cos^2x+cosx-1=0$ and then simplify to

$cosx-cos^2x=0$

Factor out the common cos(x) to get

$cosx(1-cosx)=0$ and there you have your 2 trig equations:

cos(x) = 0 and 1 - cos(x) = 0

The first one is easy enough to solve. Look on the unit circle and see where, one time around, where the cos of an angle is equal to 0. That occurs at

$x=\frac{\pi }{2},\frac{3\pi}{2}$

The second equation simplifies to

cos(x) = 1

Again, look to the unit circle and find where the cos of an angle is equal to 1. That occurs at π only.

So, in the end, your 3 solutions are

$x=\frac{\pi}{2},\pi,\frac{3\pi}{2}$

8 0

3 years ago

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Setler79 [48]

Answer:

below..

Step-by-step explanation:

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2 years ago