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Dima020 [189]
3 years ago
15

If the measures of two angles of a triangle are $2° and 78°, find the measure of the third angle.​

Mathematics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

A triangle's angles sum to 180

Step-by-step explanation:

180-2-78

=180-80

=100°

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2m=-6n-5 literal equation
Bingel [31]

Answer:

read below

Step-by-step explanation:

Alright, archtan /  

tan

−

1

(

x

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is the inverse of tangent. Tan is  

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. Like the inverse of sin, the inverse of tan is also restricted to quadrants 1 and 4.

Knowing this we are solving for the inverse of tan -1. We are basically being asked the question what angle/radian does tan(-1) equal. Using the unit circle we can see that tan(1)= pi/4.

Since the "Odds and Evens Identity" states that tan(-x) = -tan(x). Tan(-1)= -pi/4.

Knowing that tan is negative in quadrants 2 and 4. the answer is in either of those two quadrants. BUT!!! since inverse of tan is restricted to quadrants 1 and 4 we are left with the only answer -pi/4.

the answer is m=-3n-5/2 or n=2m+5/6

8 0
3 years ago
Triangle ABC has verticals A (3,0),B (9,5)and (7,-8) find the length of AC in simplest radical form
AlladinOne [14]
length =  \sqrt{(0+8)^2 + (3-7)^2} =  \sqrt{64 + 16} =  \sqrt{80} = 4\sqrt{5}


Answer: Length = 4√5 units.
4 0
4 years ago
Can 12(x^3+y^2)+6(y^2+1) be the factored form of 12x^3+18y^2+6?
Misha Larkins [42]

Answer:

YES

Step-by-step explanation:

So how you can figure this out is you do the Distributive Property.

So it will be 12x^3+12y^2+6y^2+6.

The next thing you have to do is add together the ones that are similar which is 12y^2+6y^2 to get 18y^2.

Your answer is 12x^3+18y^2+6. So YES it can be factored into that form.

4 0
3 years ago
Read 2 more answers
Which expression is equivalent to....​
Vaselesa [24]

Answer:

A^32

Step-by-step explanation:

Done

8 0
2 years ago
Read 2 more answers
Suppose Kaitlin places $6500 in an account that pays 12% interest compounded each year.
Leya [2.2K]

\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 1 year}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}

\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 1}\implies A=6500(1.12)\implies A=7280 \\\\[-0.35em] ~\dotfill

\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 2 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}

\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 2}\implies A=6500(1.12)^2\implies A=8153.6

6 0
3 years ago
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