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3 years ago

Question and answer choices in the attachment

Mathematics

1 answer:

Elden [556K]3 years ago

4 0

<h3>You have the correct answer. It is choice C) additive inverse.</h3>

The additive inverse is the idea of adding a negative number, which is the same as subtraction. Example: 2 + (-3) = 2-3

Based on the example above, if we start off with 3 and add on its additive inverse -3, then 3 + (-3) = 3-3 = 0. In general, x + (-x) = x-x = 0. So adding any number with its additive inverse gets you 0 every time.

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Extra info:

Choice A is ruled out because the commutative property of multiplication is the idea that we can multiply two numbers in any order we want. Eg: 7*8 = 56 and 8*7 = 56, so 7*8 = 8*7
Choice B is ruled out as well. The commutative property of addition is similar to the rule mentioned in choice A. But now we're adding. This rule says we can add two numbers in any order. Eg: 4+5 = 5+4 = 9
Choice D is ruled out also. The multiplicative inverse of x is 1/x. Multiplying the two together gets you 1. Keep in mind that x cannot be zero. Example: The multiplicative inverse of 99 is 1/99, since 99*(1/99) = 99/99 = 1.

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Y_Kistochka [10]

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$r=\dfrac{p}{4+\pi}$

Step-by-step explanation:

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Substitute y into the area expression:

$A=\dfrac{p-2r-\pi r}{2}\cdot 2r+\dfrac{1}{2}\pi r^2=pr-2r^2-\pi r^2+\dfrac{1}{2}\pi r^2=pr-2r^2-\dfrac{1}{2}\pi r^2.$

Find the derivative A':

$A'=p-4r-\pi r.$

Equate A' to 0:

$p-4r-\pi r=0\Rightarrow r=\dfrac{p}{4+\pi}.$

When $r$ then $A'>0$ and the function A is increasing, when $r>\dfrac{p}{4+\pi},$ then $A'$ and the function A is decreasing. This means that at point $r=\dfrac{p}{4+\pi}$ the function A takes it maximal value (the area is maximal).