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3241004551 [841]
3 years ago
7

10.5 equal or less than 10.005

Mathematics
1 answer:
maksim [4K]3 years ago
5 0
10.5 is more than 10.005
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Least to greatest<br> 4 1/2, 4.504, 4.43, 113/25
aleksklad [387]
4.43
4 1/2
4.504
113/25
8 0
3 years ago
Check if given set of numbers 6,8,10 forms pythagorean triplet or not?​
kipiarov [429]

\LARGE{ \underline{ \tt{Required \: answer:}}}

We know that,

  • Pythagoras theoram is given by : p² + b² = h²

  • In a Pythagoras triplet, the longest side represents the hypotenuse of the triangle.

Then,

We should check that, whether the sum of squares of the smaller sides will be equal to the square of the third side or not. Then, we can determine the above easily.

Sum of squares of smaller sides:

= 6² + 8²

= 36 + 64

= 100

Square of longer side:

= 10²

= 100

Sum of squares of smaller sides = Square of longer side. Hence, 6,8,10 are Pythagorean triplets.

And we are done! :D

⛱️ \large{ \blue{ \bf{FadedElla}}}

6 0
2 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
Can someone please help on this Q-Q thanks
liq [111]

Answer:

Yup it’s d

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
1.A five - layer of cake weighs half a kilogram per layer.In each layer,the cake is decorated with icing which weighs 15g.3 cand
julia-pushkina [17]

Answer:  2.815 kilograms

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

Work Shown:

1 layer = 0.5 kg

5 layers = 5*(0.5 kg) = 2.5 kg

icing = 15 g = 15/1000 = 0.015 kg

1 candle = 100 g = 100/1000 = 0.1 kg

3 candles = 3*(0.1 kg) = 0.3 kg

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

The five layers combine to 2.5 kg. On top of that we have 0.015 kg of icing, and then finally the three candles add 0.3 kg more weight.

The total weight is therefore: 2.5+0.015+0.3 = 2.815 kilograms

3 0
3 years ago
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