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Veseljchak [2.6K]
3 years ago
9

Sorry last one :Multiply. 2√⋅5√⋅11

Mathematics
2 answers:
DIA [1.3K]3 years ago
4 0
Hey! You just have to square root the 2 and 5 then multiply by 11 so the answer I came up with is 34.78505426185218

I hope I helped!

mamaluj [8]3 years ago
4 0

Answer:

√110

Step-by-step explanation:

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codys bikes rents bikes for $18 plus $5 per hour. mofor paid $48 to rent a bike. For how many hours did he rent the bike?
DerKrebs [107]
First of all, to find out the amount paid for hours in total,
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3 0
3 years ago
Read 2 more answers
Plz help I will mark as the brainlist
tensa zangetsu [6.8K]

Answer:

1/25

Step-by-step explanation:

We can see that in this pattern, the denominator and the numerator of the next fraction cancel out; they're the same.

(1/2) x (2/3) x (3/4) = 1/4 (do you see it?)

The only things that do not cancel out are the numerator of the first fraction and the denominator of the last. So, in the end, the answer will be 1/25.

Hope this helps!

3 0
3 years ago
Help me with this plsss 40 pointsss
lara [203]

Answer:

Step-by-step explanation:

x(3x+2)=16

3x²+2x-16=0

3x²+8x-6x-16=0

x(3x+8)-2(3x+8)=0

(3x+8)(x-2)=0

either 3x+8=0

x=-8/3(rejected as width can't be negative)

x-2=0

x=2

so width=2

length=3×2+2=6+2=8

7 0
3 years ago
Find the exact value of tan(165°) using a difference of two angles
Katyanochek1 [597]

Answer:  -2+\sqrt{3}

=========================================================

Work Shown:

Apply the following trig identity

\tan(A - B) = \frac{\tan(A)-\tan(B)}{1+\tan(A)*\tan(B)}\\\\\tan(225 - 60) = \frac{\tan(225)-\tan(60)}{1+\tan(225)*\tan(60)}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+1*\sqrt{3}}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\

Now let's rationalize the denominator

\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\\tan(165) = \frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}\\\\\tan(165) = \frac{(1-\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{(1)^2-2*1*\sqrt{3}+(\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{1-2\sqrt{3}+3}{1-3}\\\\\tan(165) = \frac{4-2\sqrt{3}}{-2}\\\\\tan(165) = -2+\sqrt{3}\\\\

----------------------

As confirmation, you can use the idea that if x = y, then x-y = 0. We'll have x = tan(165) and y = -2+sqrt(3). When computing x-y, your calculator should get fairly close to 0, if not get 0 itself.

Or you can note how

\tan(165) \approx -0.267949\\\\-2+\sqrt{3} \approx -0.267949

which helps us see that they are the same thing.

Further confirmation comes from WolframAlpha (see attached image). They decided to write the answer as \sqrt{3}-2 but it's the same as above.

5 0
3 years ago
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