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Advocard [28]
4 years ago
15

14 - 10 1/6 . PLEASE ANSWER I WILL GIVE BRAINLIEST

Mathematics
2 answers:
BaLLatris [955]4 years ago
4 0

Answer:

3.8333

Step-by-step explanation:

I hope this helps

Tems11 [23]4 years ago
3 0

Answer: 3.8333333

Step-by-step explanation:

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One leg of a right triangle is five and it's hypotenuse is 10. What is the size of its smallest angle?
galina1969 [7]
I hope this helps you



30 degree = k


60 degree = k square root of 3


90 degree = 2k


hypothenus =10 =>90 degree


one leg = 5 => 30 degree
4 0
3 years ago
Read 2 more answers
what is the answer to 3x^2+5x+1 PPPPPLLLLEEEEAAAASSSSEEE PLEASE HHHHEEEELLLPPP ME QUICK!!!!!!!!!!!!!!!!! SHOW WORK PLEASE
aleksandr82 [10.1K]
3x to the second power plus 5x plus 1
6 0
3 years ago
I will give you a good amount of points if you answer this correct​
Colt1911 [192]

Answer:

I think the answer is B. The equation has infinitely many solutions.

Step-by-step explanation:

4(3x + 4) = 15x + 12 - 3x + 4

*group like terms so 4(3x+4) = 15x - 3x + 12 + 4

*add similar elements 15x - 3x = 12x

*add the numbers 12 + 4 = 16

*Expand to 12x + 16 - 16 = 12x + 16 - 16

*You simplify 12x = 12x

*Subtract 12x from both sides

*Then Simplify which is 0

Which means Both sides are equal to 0

True for all X

5 0
3 years ago
Find the equation of the line through point (2,−4) and parallel to −6x+2y=4. Use a forward slash (i.e. "/") for fractions (e.g.
chubhunter [2.5K]

Answer:

y = 3x - 10

Step-by-step explanation:

Step 1: Rewrite 1st equation

2y = 6x + 4

y = 3x + 4

Step 2: Find 2nd equation

y = 3x + b

-4 = 3(2) + b

-4 = 6 + b

b = -10

Step 3: Rewrite 2nd equation

y = 3x - 10

And we have your parallel equation!

5 0
3 years ago
An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu
exis [7]
Take the derivative with respect to t
- w \sin(wt) + \sqrt{8} w cos(wt)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
0 = -w \sin(wt) + \sqrt{8} w cos(wt)
divide by w
0 =- \sin(wt) + \sqrt{8} cos(wt)
we add sin(wt) to both sides

\sin(wt)= \sqrt{8} cos(wt)
divide both sides by cos(wt)
\frac{sin(wt)}{cos(wt)}= \sqrt{8}   \\  \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\  \\ wt=arctan(2 \sqrt{2)} OR\\ wt=arctan( { \frac{1}{ \sqrt{2} } )
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

I=cos(2n \pi -2arctan( \sqrt{2} ))
since 2npi is just the period of cos
cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 

substituting our second soultion we get
I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))
since 2npi is the period
I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}
so the maximum value =\frac{1}{3}
minimum value =- \frac{1}{3}


4 0
3 years ago
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