Ask question

Ask question

All categories

3 years ago

11

The hypotenuse and one of the legs of a right triangle form an angle that has a cosine of 2 √2 .

2 answers:

Tems11 [23]3 years ago

7 0

Quick answer I don't think this has an answer.

If you take the cos-1(2 sqrt(2)) your calculator should have a fit. Let's check that out. Mine certainly does. So there is something wrong with the question. If there is something to add in please do it and I will it least put an answer in the comments. As it stands, nothing will work.

If you put your calculator in radians, you will get an answer but it will not be anything resembling the choices you've listed.

If you meant sqrt(2) / 2 that would give 45o. Put it in your calculator like this 2 ^ 0.5 divided by 2 = 0.707
Cos - 1 (0.707) = 45

IceJOKER [234]3 years ago

5 0

The answer is B. 45 degrees

You might be interested in

Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

<u>I will try to give as many details as possible. </u>

First of all, I just would like to say:

$\text{Use } \LaTeX !$

Texting in Latex is much more clear and depending on the question, just writing down without it may be confusing or ambiguous. Be together with Latex! （＊＾Ｕ＾）人（≧Ｖ≦＊）/

$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

As

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

4 0

3 years ago

Mumz [18]

If you use/make a number line, then you will see that the answer is -18 or a. When using a number line, negative = ← and positive = →.

6 0

3 years ago

Hello pls help besties (o˘◡˘o)

goldenfox [79]

Hello bestie I cannot help you

8 0

3 years ago

Read 2 more answers

CONVERT 2930 TO BASE FIVE

prohojiy [21]

Hello

The answer is <span>43210<span>5

Have a nice day</span></span>

5 0

3 years ago

Which has a sum = to 2 3/8 +3 4/8?

monitta

B) 19/8 + 28/8

have a great day!

3 0

4 years ago