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zhenek [66]
3 years ago
13

Find the area of the rectangle. 20 and 9.

Mathematics
1 answer:
weeeeeb [17]3 years ago
3 0

Answer:

180

Step-by-step explanation:

20 x 9 = 180

Let me know if you need more help!

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a student claims that 1,2,3, and 4 are the zeros of a cubic polynomial function. explain why the student is mistaken
Alex_Xolod [135]
A cubic polynomial function has 3 zeroes, which is why it is cubic. The student says there are 4 zeroes, which is not possible.
7 0
3 years ago
Help me find the volume of the composite figure
Butoxors [25]

Answer:

V = 97.911in^3

Step-by-step explanation:

Given

Shapes: Cube and Cone

Required

Determine the volume

First we calculate the volume of the cube

V_1 = l^3

Where

l = side length = 5.1

V_1 = 5.1^3

V_1 = 132.651

Next, calculate the volume of the cone using:

V_2 = \frac{\pi r^2h}{3}

Where

h = 5.1

r = \frac{1}{2} * Diameter

r = \frac{1}{2} * 5.1

r = 2.55

So, we have:

V_2 = \frac{\pi r^2h}{3}

V_2 = \frac{\pi * 2.55^2 * 5.1}{3}

V_2 = \frac{22 * 2.55^2 * 5.1}{7*3}

V_2 = \frac{729.5805}{21}

V_2 = 34.74

The volume of the figure is:

V=V_1 - V_2

V_1 = 132.651

V = 132.651 - 34.74

V = 97.911in^3

8 0
3 years ago
HELP PLEASE IM HORRIBLE WILL GIVE A BRAIN LIST IF RIGHT
Bogdan [553]

Answer:

you did not even put anything except "HELP PLEASE IM HORRIBLE WILL GIVE BRAIN LIST IF RIGHT"

7 0
3 years ago
Read 2 more answers
The formula a=46c gives the floor area (a) in square meters that can be wired using (c) circuits
antoniya [11.8K]
 a = 46c....for c
a/46 = c

322 = 46c
322/46 = c
7 = c.....7 circuits are required
6 0
3 years ago
How do I find the Derivative of a Function where the x is a number?
lisov135 [29]

Given a function <em>g(x)</em>, its derivative, if it exists, is equal to the limit

g'(x) = \displaystyle\lim_{h\to0}\frac{g(x+h)-g(x)}h

The limit is some expression that is itself a function of <em>x</em>. Then the derivative of <em>g(x)</em> at <em>x</em> = 1 is obtained by just plugging <em>x</em> = 1. In other words, find <em>g'(x)</em> - and this can be done with or without taking a limit - then evaluate <em>g'</em> (1).

Alternatively, you can directly find the derivative at a point by computing the limit

g'(1) = \displaystyle\lim_{h\to0}\frac{g(1+h)-g(1)}h

But this is essentially the same as the first method, we're just replacing <em>x</em> with 1.

Yet another way is to compute the limit

g'(1) = \displaystyle\lim_{x\to1}\frac{g(x)-g(1)}{x-1}

but this is really the same limit with <em>h</em> = <em>x</em> - 1.

You do not compute <em>g</em> (1) first, because as you say, that's just a constant, so its derivative is zero. But you're not concerned with the derivative of some <em>number</em>, you care about the derivative of a function that depends on a <em>variable.</em>

8 0
3 years ago
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