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3 years ago

I'd appreciate it if anyone could help ASAP. I'm giving Brainliest!

Mathematics

1 answer:

andre [41]3 years ago

7 0

Answer:

(1, 0.25)

Step-by-step explanation:

To find when f(x)=g(x), then look for a value that is the same for both functions in the table. 0.25 occurs twice for the same x value. This is when they are equal. The solution is (1, 0.25).

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What’s the area of a triangular pyramid

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Answer:

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Step-by-step explanation:

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3 years ago

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Without calculating, which has a bigger volume. A cube that has a length, width, and height of 18 m. Or a sphere with a radius o

Evgesh-ka [11]

Weird. A period appears above this... huh.

Answer:

[Th]e cube has a greater value.

Step-by-step explanation:

What the word problem really wants us to get [is ]the question of 'Which is greater, A=6a^2 when 'a' [is] 18 or A=4 $\pi$ r^2 when r = 9? And here's how to solve that.

Starting with the[ c]ube we have A=6a^2. A bit t[o]o simple, right?

A=6(18)^2 Substitute numbers.

A=6(324) Solve ex[p]onents.

A=1944 Mult[i]ply.

So w[e] know that the cube is 1944 meters cube[d ] in area. But what about the more [f]ormidable sphere? Fo[r] it we need a slightly m[o]re co[m]plicated formula, A=4 $\pi$ r^2. However, instead of using the real pi I will be rounding to 3.14, since we have no calculator so anything more would take way too long and fry your[ bra]in.

A=4(3.14)(9)^2 Subst[i]tute numbers.

A=4(3.14)(81) Solve expone[n]ts.

A=12.56(81) Multip[ly].

A=1017.36 Multiply again[.]

Now, since I'm sure all of us can count, we know that 1944 is greater than 1017.36. Or in other words, the cube is bigger than the sphere.

And PLEASE don't copy this guys. Make your own iteration. Change it up!

3 0

3 years ago

Suppose that a factory manufactures only tables and chairs and that the profit on one chair is $15 and on the table is $20. Each

Leya [2.2K]

Answer:

The data we have is:

Profit per chair = $15

Profit per table = $20

Chair needs: 1 large pice, 2 small pieces.

Table needs: 2 large pieces, 2 small pieces.

You have 6 large pieces and 8 small pieces, and you want to maximize the profit.

Let's define:

T = # of tables

C = # of chairs.

Total profit:

P = C*$15 + M*$20

The number of chairs that you can make is given by:

0 ≤ C ≤ 4

If you make 4 chairs, the profit is:

P = 4*$15 = $60

And there will be 2 large pieces left.

Now for the tables.

0 ≤ T ≤ 3

If you make 4 tables, the profit will be:

P = 3*$20 = $60 (same as before)

And there will be two small pieces left.

Then we want to use a medium number of tables and chairs, we can not use the maximum for any of those, the first thing we should try is:

we ideally would want to use all the pieces that we have.

8 small, 6 large:

Then we can make:

2 tables: we use 4 large pieces and 4 small pieces.

T = 2

And we have left 2 large pieces and 4 small pieces.

The leftover is enough to make two chairs.

C = 2.

In this case, where we used all our materials, the profit will be:

P = 2*$15 + 2*$20 = $30 + $40 = $70

The profit is maximized when we make 2 tables and two chairs.

5 0

3 years ago

99 points

Scrat [10]

$l \ equal \ 3m \\\\ therefore, 6 *L^2=6*3^2 = (6*9)$

$((6*9)) = 54m^2$

<span>27 m²
30 m²
36 m²
54 m²</span>

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