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egoroff_w [7]
3 years ago
15

Write the Riemann sum to find the area under the graph of the function f(x) = x^2 from x = 1 to x = 5.

Mathematics
2 answers:
prohojiy [21]3 years ago
8 0
\lim_{n \to \infty}∑ -- agggg

The 2nd one, but get me a minute to write up the explaination/equation I can't do it with the symbol brainly provides



makkiz [27]3 years ago
6 0
I think the first one or the thrid one 
hope this help
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A=1,200×(1−0.25)^(3)
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12x>9 (2x-3)-15 solve for x and show steps please.
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12x > 9(2x-3)-15

distribute the 9

12x > 18x-27 -15

combine like terms on the right side

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Help me please!!!!!!!
Cerrena [4.2K]

Answer:

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Step-by-step explanation:

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You have a wire that is 20 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The o
Aleksandr [31]

Answer:

Therefore the circumference of the circle is =\frac{20\pi}{4+\pi}

Step-by-step explanation:

Let the side of the square be s

and the radius of the circle be r

The perimeter of the square is = 4s

The circumference of the circle is =2πr

Given that the length of the wire is 20 cm.

According to the problem,

4s + 2πr =20

⇒2s+πr =10

\Rightarrow s=\frac{10-\pi r}{2}

The area of the circle is = πr²

The area of the square is = s²

A represent the total area of the square and circle.

A=πr²+s²

Putting the value of s

A=\pi r^2+ (\frac{10-\pi r}{2})^2

\Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2

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For maximum or minimum \frac{dA}{dr}=0

Differentiating with respect to r

\frac{dA}{dr}= \frac{2\pi r(4+\pi)}{4} -5\pi

Again differentiating with respect to r

\frac{d^2A}{dr^2}=\frac{2\pi (4+\pi)}{4}    > 0

For maximum or minimum

\frac{dA}{dr}=0

\Rightarrow \frac{2\pi r(4+\pi)}{4} -5\pi=0

\Rightarrow r = \frac{10\pi }{\pi(4+\pi)}

\Rightarrow r=\frac{10}{4+\pi}

\frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0

Therefore at r=\frac{10}{4+\pi}  , A is minimum.

Therefore the circumference of the circle is

=2 \pi \frac{10}{4+\pi}

=\frac{20\pi}{4+\pi}

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Answer:

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