All Answers 358090 Answers

2+2=4

4 is your answer.

Really hard.

How to solve 14+12*x when x = 3

14+12*x

Subsitute x when x = 3.

14+12*3

Use the order of operations to solve the problem.

Multiplication

Division

Addition

Subtraction

So, we first mulltiply.

12*3=36

Finally, to get the answer, we have to have add 36 and 14.

36+14=50 ANSWER

So, the answer is 50.

CPhill, I am a bit confused on this one. May you please explain it to me?

I had gotten up to 70 but I did not know what to do after lol thank you cphill

The two radii are equal sides of an isosceles triangle. So the angles opposite those sides (A and B) are equal. Thus, angle A + angle B + 110 = 180. So, angle A + angle A + 110 = 180. Subtract 110 from each side. Thus angle A + angle A = 180-110. So 2*angle A  = 70. Divide by 2 on both sides. Then, angle A = (70)/2 = 35 degrees.

Use the "cos" button that is the third key in the 1st column of the scientific calculator on this website.

For the second question, relating the graph to f(x) = cosx with interval -2pi <= x <= 2pi:

Notice that for f(x) = cosx, if x = 0, then f(0) = 1. But, in this graph at cos(0), the value is 2 which suggests that the graph had to have been shifted up one unit. This graph is f(x) = cosx + 1 = cos(0) + 1 = 1 + 1 = 2.

Simplifying, we have...

x - 6 - 6x + 36 = 0 

-5x + 30 = 0     subtract 30 from both sides

-5x = -30     divide both sides by -5

x = 6

Very nice explanation, Aziz...!!!

Here is another way to look at this, analytically (if not particularly "mathematically")...Note that, when lyl =0,  lxl cannot be larger than 1. Thus, x cannot be more than +1 or less than -1. So, plot the two points on the graph (1,0) and (-1,0). Likewise, when lxl = 0, lyl cannot be more than 1 or less than -1. Then, also plot the two points (0,1) and (0, -1) on the graph. Then, let all four points be the vertices of a square. The sides of this square are the "bounds" of our graph. Note that every point lying on any of the four sides satisfies lxl + lyl = 1, and any point lying "outside" these bounds, doesn't. (They make the equation > 1).

This gives us a way to graph any equation of the form lxl + lyl = a....where a is some positive real number.

Does that help??

lxl + lyl = 1

if x = 0, then y = 1,

and if y = 0, then x = 1.

So (0,1) and (1,0) are definitely points.

Also, we can also do the same for:

if x = 0, then y= -1 as well since we have the absolute value sign,

and if y = 0, then x = -1 as well since we have the absolute sign.

So, we now have 4 main points (0,1), (1,0), (0, -1), & (-1,0). 

Now, we know x or y cannot be greater than 1 since lxl + lyl must be equal to 1 [if x = 2, then there would be no value for y that would satisfy the equation].

Furthermore, we notice that 0.5 + 0.5 also equals 1 so we get more points: (0.5, 0.5), (-0.5, 0.5), (-0.5, -0.5), & (0.5, -0.5).

We now have 8 points and if you connect them together, you get a square.

So, these 8 points create a square with sides of sqrt(2).

lxl + lyl = 1

We need to get rid of the two absolute value signs from x and y:

Since we have two variables and each absolute value has 2 possiblities, we would end up with a total of 4 equations [abs(x) = x / -x & abs(y) = y / -y]:

x + y = 1, x - y = 1, -x + y = 1, & -x - y = 1

1) x + y = 1 --> y = -x + 1 --> Notice, we took positive x and y, so we are interested in values: x and y are >= to 0. If you graph this, you get a line that is only in the 1st quadrant.

2) x - y = 1 --> y = x - 1 --> We took positive x and negative y, so we are interested in values: x is >= 0 and y < 0. If you graph this, you get a line that is only in the 4th quadrant.

3) -x + y = 1 --> y = x + 1 --> Follow the pattern...negative x and positive y means the 2nd quadrant.

4) -x - y = 1 --> y = -x - 1 --> Follow the pattern...negative x and negative y means the 3rd quadrant.

Notice that each of the 4 lines has a slope of 1 or -1.

Now, if you graph each of these, you would find that it looks like the graph Phil posted of a square that has sides of sqrt(2).

but how you draw it .. and is the drawing the only way ! 

The area is a square with a side of  √2....thus the area = ( √2)²  = 2

A graph might help........

"The area of the region enclosed by |x| + |y| = 1 is:"

Have fun and enjoy your time there, Rosala!

xvxvxv, I can't see the first question. Can you tell me what the first question is so I can solve it?

The answer is trillion one. And CPhill, I found some of your glasses in my house! I will give you back your glasses!

Hope you got your glasses back!

Key it in like this......

.105-1.96*sqrt(.105*.895/258)

2+2=4 

4 is the answer.

Thanks for giving my points and everybody's points back, CPhill! I appreciate that!

0/0=0. If you divide or multiply 0, the answer is ALWAYS 0.

Good luck! See yah, rosala!

The Queen seemeth to be off her rocker.....er, I mean throne......She claimeth that a "real" number was unspecified, yet, was not her answer a "real" number??? Further, this disputes M'Lady's assertion that a woman winneth the argument >95% of the time....Sir Cphill has given an answer which hath more power than the Queen's doth have.....thus, Sir Cphill hath won 100% of the time in this argument.....

Also, two things come, alphabetically, after a "Trilllion"....."Trilobite" (a Fossil) and "Tyrannosaurus" (a Dinosaur). But, the Fossil precedeth the Dinosaur....thus, the Fossil possesseth more precedence than the Dinosaur in all things.......

The Queen may now returneth to her rocker......er, throne......

Your Humble Answerer,

Sir CPhill