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3 years ago

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What is another way to write (3x1000)+(3x10)+(3x1/100)+(3x1/1000

1 answer:

Paha777 [63]3 years ago

3 0

That'd be 3000 + 30 + 0.03 + 0.003, or 3030.033

hope this helped :)
alisa202

plzz tell me if I was wrong

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Analyze the data sets below. Which of the following statements are true?

Fed [463]

I do not see the statements.

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3 years ago

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Compare 4.5x10^6 and 2.1x10^8 explain how you can compare them

Roman55 [17]

4.5 x 10^6 = 4,500,000
2.1 x 10^8 = 210,000,000

Therefore, 4.5 x 10^6 < 2.1 x 10^8

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Triangle KML and triangle PRQ are shown in the figure belowK404008 cms4 cms10 cmsOL3 cms6 cmsRM17) Are these triangles similar?1

s2008m [1.1K]

Two triangles are similar if the only difference between them is the size, this means that their internal angles must be the same. If we look at the picture the first triangle has one angle equal to 40 degrees, one equal to 80 degrees and the third one is unkown (x). The second triangle has one angle equal to 40 degrees, one equal to 60 degrees and the third one is unkown (y). The sum of the internal angles of a triangle must be equal to 180 degrees, with this information we can find the values of the missing angles. We have:

$\begin{gathered} 40+80+x=180 \\ 120+x\text{ = 180} \\ x\text{ = 180-120=60} \end{gathered}$ $\begin{gathered} 40+60+y=180 \\ 100+y=180 \\ y=180-100=80 \end{gathered}$

Therefore the internal angles of the first triangle are (40,80,60) and the angles of the second triangle are (40,80,60) as well, therefore they are similar.

Two triangles are congruent if they have sides with the same length. Which is not the case, because the sides of one triangle is (8, 10, 6) while the other is (4,3 and unkown). Therefore they are not congruent.

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1 year ago

What value of z makes the equation true?<br> 5/6=z/12<br> A. 6<br> B. 10<br> C. 15<br> D. 5

vazorg [7]

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3 years ago

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Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

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3 years ago