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3 years ago

Find the lenght of the missing side:

Mathematics

2 answers:

Reika [66]3 years ago

8 0

Answer:

15.73

Step-by-step explanation:

cos(x) = adjacent leg / hypotenuse

cos(70) = x / 46

0.342 = x / 46

0.342 x 46 = x

x = 15.73

ollegr [7]3 years ago

7 0

Answer:

x=15.73292

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

cos theta = adj/ hyp

cos 70 = x / 46

46 cos 70 =x

x=15.73292

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Which of the following inequalities matches the graph? graph of an inequality with a solid line through the points (1, 4) and (2

Ganezh [65]

Is 5x +y greater than or equal to 1

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3 years ago

Read 2 more answers

Factor completely<br> 2c^3-22c^2+48c

Bingel [31]

Answer:

2c(c - 3)(c - 8)

Step-by-step explanation:

Hello!

Factor:

2c³ - 22c² + 48c
2c(c² - 11c + 24)

Think: What two numbers multiply to 24, and add up to -11?

Answer: -8 and -3

Split up the -11c into -8c and -3c:

2c(c² - 8c - 3c + 24) (factor each pair)
2c( c(c - 8) - 3(c - 8))
2c(c - 3)(c - 8)

7 0

3 years ago

Which is the correct answer?? I have 5 mins left to finish this pls help

NeX [460]

Answer:

25.3

Step-by-step explanation:

I just put it in my calculator.

5 0

3 years ago

Teje ratio of adulta to children attedening a new exhibit at the museum was found 8:5 based on this ratio if 390 people attended

Ratling [72]

8x - adults
5x - children

$8x+5x=390\\
13x=390\\
x=30\\\\
5x=5\cdot30=150$

8 0

3 years ago

How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%

blondinia [14]

The trick is to exploit the difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}$

So we have

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2$

But √12 = √(3•4) = 2√3, so

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

7 0

3 years ago