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2 years ago

What is a plane in mathematical terms

Mathematics

1 answer:

Whitepunk [10]2 years ago

5 0

Answer:

Step-by-step explanation:

A plane is a flat, two-dimensional surface that extends infinitely far.

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Step-by-step explanation:

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2 years ago

Jamie and Owen joined a monthly walking challenge at their youth center. Jamie has already walked 120 minutes and plans to walk

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The diagonals of rectangle nopq intersect at point r. if oq=2(x+3) and pr=3x-5, solve for x.

OverLord2011 [107]

The diagonals of a rectangle are always congruent. We have the whole length of OQ to be 2x + 6, and we have only half of NP. Therefore, OQ = 2(PR). 2x+6=2(3x-5). 2x+6 = 6x-10. 16 = 4x so x = 4. That means that each diagonal is 14 units long.

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2 years ago

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-

Ostrovityanka [42]

Answer:

2.25

Step-by-step explanation:

The computation of the number c that satisfied is shown below:

Given that

$f(x) = \sqrt{x}$

Interval = (0,9)

According to the Rolle's mean value theorem,

If f(x) is continuous in {a,b) and it is distinct also

And, f(a) ≠ f(b) so its existance should be at least one value

i.e

$f^i(c) = \frac{f(b) - f(a)}{b -a }$

After this,

$f(x) = \sqrt{x} \\\\ f^i(x) = \frac{1}{2}x ^{\frac{1}{2} - 1} \\\\ = \frac{1}{2}x ^{\frac{-1}{2}$

$f^i(x) = \frac{1}{{2}\sqrt{x} } = f^i(c) = \frac{1}{{2}\sqrt{c} } \\\\\a = 0, f (a) = f(o) = \sqrt{0} = 0 \\\\\ b = 9 , f (b) = f(a) = \sqrt{9} = 3\\$

After this,

Put the values of a and b to the above equation

$f^i(c) = \frac{f(b) - f(a)}{b - a} \\\\ \frac{1}{{2}\sqrt{c} } = \frac{3 -0}{9-0} \\\\ \frac{1}{\sqrt[2]{c} } = \frac{3}{9} \\\\ \frac{1}{\sqrt[2]{c} } = \frac{1}{3} \\\\ \sqrt[2]{c} = 3\\\\\sqrt{c} = \frac{3}{2} \\\\ c = \frac{9}{4}$

= 2.25

4 0

2 years ago