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Question in LaTeX format. welcom

If 1/x < 2 and 1/x > -3, then find all possible values of x.

Hello Guest!

\(\frac{1}{x_1}<2\\ x_1>\frac{1}{2}\)

\(\frac{1}{x_2}>(-3)\\ x_2<(-\frac{1}{3})\)

\(x\in \mathbb R\ |x<(-\frac{1}{3})\ and\ x>\frac{1}{2}\)|

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The scenario can be modeled by the function f(x)=80(3)^x/5 , where  f(x) represents the population of bacteria x hours after the bacteria are placed in the Petri dish. [TRUE or FALSE ?] - THIS IS TRUE.

After 13 hours, there are approximately 1592 bacteria. [TRUE or FALSE ?]-THIS IS FALSE!. THE NUMBER SHOULD BE: 1392 NOT 1592!.

Think all the women together as one object.

Now how would you arrange that now?

Let M be men, and W be women.

There are now 4 Ms and one W.

Solve 2^{2x} - 8 * 2^x + 12 = 0.

Hello guest!

\(2^{2x} - 8 * 2^x + 12 = 0. \\ u=2^x\\ u^2-8u+12=0\\ u=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ u=4\pm\sqrt{16-12}\)

\(u_1=6\\ u_2=2\)

\(2^{x_1}=6\\ x_1\cdot ln2=ln6\\ x_1=\frac{ln6}{ln2}\)

\(x_1=2.5849625\)

\(2^{x_2}=2\\ 2^{x_2}=2^1 \)

\(x_2=1\)

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Red area = pi x r^2 = pi (1^2) = pi

Green area = pi r^2  - the red area = 9 pi - pi = 8 pi

Blue area = pi r^2 - green - red areas = 25pi - 8pi - pi = 16 pi

Total area is   pi r^2 = 25 pi                        (16pi + 8pi + pi)

green area is 8 pi

   8pi/25pi = 8/25

4 cooks

4 hours

4 cakes

4 cakes / (4 cooks X 4 hours)  =   4 cakes / 16 cook-hours = 1 /4  cake/cook-hours

8 cakes /  1/4 cake/cook-hour   = 32 cook hours

32 cook-hour / 8 cooks = 4 hours

You mean " Divisible EVENLY without aremainder" ?

(1): An odd number is never divisible by an even number............True.e.g. 15 / 2 =7.5 leaves remainder.
(2): An odd number is never divisible by another odd number...  False.e.g. 21 / 7 = 3
(3): An even number is never divisible by an odd number............False.e.g.  24 / 3 = 8
(4): An even number is never divisible by another even number..False.e.g.  24 /8 = 3

The smallest n would be 4 itself, because:

GCD{4, 16, 20} = 4, or:

GCD{8, 16, 20} = 4, or:

GCD{12, 16, 20} = 4 and so on......

Let $z$ be a complex number such that z^2 + |z|^2 = 3 - 5i.Find |z|^2.

\(Let \;\;z=a+bi   \\ z^2= a^2-b^2+2abi\\ |z|^2=a^2+b^2\\ so a^2-b^2+2abi + a^2+b^2 = 3-5i \\ 2a^2+2abi  = 3-5i \\ 2a^2=3 \qquad  2ab=-5\\ a^2=1.5 \qquad 2a^2b^2=25\\ \qquad \qquad \quad 2*1.5b^2=25\\ \qquad \qquad \quad b^2=\frac{25}{3}\\ so\\ |z|^2=\frac{3}{2}+\frac{25}{3}=\frac{106}{6}=17\frac{2}{3}\)

As always, you must check this answer.

x^2 + 0x + 0 (simplified to just x^2) only has one root of 0,0 if that counts

No cake takes 4 hours in the oven to bake. Therefore, each cake takes one hour in the oven. Since you have 8 cakes, it will take 8 hours. Please correct me if I'm wrong other members on 2.0calc. 

Thanks Alan, impressive as always :)

idk  Can you give me more info on what you want the u substitution to be?

Solve for x:
sqrt(x + 4) = x - 2

Raise both sides to the power of two:
x + 4 = (x - 2)^2

Expand out terms of the right hand side:
x + 4 = x^2 - 4 x + 4

Subtract x^2 - 4 x + 4 from both sides:
5 x - x^2 = 0

Factor x and constant terms from the left hand side:
-x (x - 5) = 0

Multiply both sides by -1:
x (x - 5) = 0

Split into two equations:
x - 5 = 0 or x = 0

Add 5 to both sides:
x = 5 or x = 0

So, x = 0 discard.  Or x = 5

ALL the 2-digit numbers from 34 -99 = 66 numbers, when added 3 at a time, will make a 3-digit number. So, theoretically, you could have as many as 66 C 3 =45,760 3-digit numbers!!! In addition, if you are allowed to, you can take any of the 66 numbers and multiply it by 3, which is the same as adding them 3 times such as: 34 + 34 + 34 =102. However, since there are only 900 3-digit numbers between 100 and 999, most of the above numbers will be duplicates many times over.

I can only think of n = 1

112 mod [6, 9, 15, 18] = 4

You have at least 3:

15317  30617  61217.

The smallest is of course =15,317.

224 = 2^5  x  7.  All we need to do here is to multiply  224 by 2 x 7 =14, which makes the factors =2^6 x 7^2=3,136 which is a perfect square. So, the smallest n = 14.

Square root of 3,136 = 56.

In three different ways as follows:

1  - 402  to 406 = 2020
2  - 249  to 256 = 2020
3  - 31  to 70 = 2020

First, we know that segment PS is 24 units long. We will want to find QS in order to find RS. So, let's find QS

We know that PQ is half as long as QS, meaning 2 PQ = QS

We can take this as PQ + QS = PS -> PQ + 2PQ = PS -> 3PQ = PS

PS is 24 units.

3PQ=24

Divide by 3.

PQ=8

Finding PQ, we can now figure out the answer for 2PQ = QS.

2(8)=QS

16=QS

QS is 16 units long. 

In order to find RS, we need to create a similar equation as finding QS

QS= RS + QR

QR=3RS

Substitute 3RS for QR in the first equation. Substitute 16 for QS in the first equation

16 = RS + 3RS

16 = 4RS

RS = 4

So, we now know that RS is 4 units long.