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2 years ago

Marvin wrote down all of the numbers from 1 to 100. How many times did the digit 8 appear in Marvins's list?

Mathematics

1 answer:

Nata [24]2 years ago

6 0

Answer:

Originally Answered: How many times does 8 occur in 1 to 100? 20 8's occur from 1 to 100. 8, 18, 28, 38, 48, 58, 68, 78, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89 and 98.

Step-by-step explaintion: i am 5 million iq brainlist pls

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Rectangle A has a length of 2x + 6 and a width of 3x. Rectangle B has a length of x + 2 and an area of 12 square units greater t

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So here is how you solve for the answer.
Firstly, you solve for the Area of Rectangle A.
The formula for Area is Length x width.
So A = (2x + 6)(3x) and the result is: 6x^2 + 18x
Now, let y be the width of rectangle B.
<span>(x+2) (y) = 6x^2 + 18x + 12
(x+2) y = 6(x+1)(x+2)
y = 6(x+1)
</span>So the final answer would be width is 6x + 6. The answer is the third option. Hope this answer helps.

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Colt1911 [192]

See the attached picture.

Create an equation for the volume of the box, find the zeroes, and sketch the graph of the function.

The resulting box has a volume

$V(x)=x(8-x)(12-x)=x^3-20x^2+96x$

because the volume of a box is the product of its width $(12-x)$ , length $(8-x)$ , and height $(x)$ .

find the zeroes

You know right away from the factored form of $V(x)$ that the zeroes are $x=0,8,12$ . (zero product property)

sketch the graph of the function

Easy to plot by hand. You know the zeroes, and you can check the sign of $V(x)$ for any values of $x$ between these zeros to get an idea of what the graph of $V(x)$ looks like. See the second attached picture.

Here's what I mean by "check the sign" in case you don't follow. We know $V(x)=0$ when $x=0$ and $x=8$ . So we pick some value of $x$ between them, say $x=1$ , and find that

$V(1)=1(8-1)(12-1)=7\cdot11=77$

which is positive, so $V(x)$ will be positive for any other $x$ between 0 and 8. Similarly we would find that $V(x)$ for $x$ between 8 and 12, and so on.

What is the size of the cutout he needs to make so that he can fit the most marbles in the box?

It's impossible to answer this without knowing the volume of each marble...

If Thomas wants a volume of 12 cubic inches, what size does the cutout need to be?

Thomas wants $V(x)=12$ , so you solve

$x^3-20x^2+96x=12$

While this is possible to do by hand, the procedure is tedious (look up "solving the cubic equation"). With a calculator, you'd find three approximate solutions

$x\approx0.1284$