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3 years ago

13

What is the answer to the equation: 8+10

2 answers:

Greeley [361]3 years ago

4 0

Answer:

18 the answer

hope it helps

alina1380 [7]3 years ago

3 0

Answer:

18, the answer to 8+10 is 18, like you learn in 1st grade

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Third grade 1 - 4/7 =?

Hitman42 [59]

Answer:

3/7

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0.4285714286

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3 years ago

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The size of a rectangular photo frame is 12 11/16cm by 8 3/4cm. Which of the following gives the product of these dimensions rou

Alexxx [7]

Answer:

D

Step-by-step explanation:

the answer can't be A or C because we know that 8 3/4 rounds to 9

so we have to figure out if 12 11/16 is closer to 12 or 13

anything above 8/16 should round up; therefore, 11/16 should round the 12 to be 13

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13 / 1/4 jrjhdbfhfnfndnsnd

gtnhenbr [62]

The answer would be 52

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3 years ago

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Kite GHIJ has sides GH, HI, IJ and JG. What are the diagonals of this figure? Select all that apply.

juin [17]

Answer:

GI or HJ

Step-by-step explanation:

Diagonals connect non-adjacent vertices.

That is, the diagonal with G as an endpoint will not connect to vertices H or J, but will connect to vertex I. Likewise the diagonal with H as one end will have J as the other end. A quadrilateral has only two (2) diagonals. Of course, each can be named two ways:

GI or IG

HJ or JH

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3 years ago

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 $-1\le \sin \beta \le1$

4 0

3 years ago