All Answers 358095 Answers

i love it!

For some people it is hard, and for some people it comes easy. I'd say just try your hardest even if it is difficult. You can really do anything you put your mind to.

"Mathematics is the study of topics such as quantity (numbers), structure, space, and change."

Your going to have to specify...there are many ways to find the slope when you are given different information.

To trun a decimal into a percentage, move the decimal over two to the right, and add a percent sign.

$$0.45 = 45\%$$

It should be. Let us know if you find an error or try cross-referencing it with another calculator.

7x² - 36x + 12

Since the last term is positive, the signs must be both positive or both negative.

Since the middle term is negative, the signs must be both negative.

There is only one set of factors for 7:  1 and 7.

However, there are many factors for 12:  12 and 1,  6 and 2,  4 and 3 (and they have to be checked both ways).

So the possibilities are:  

     (7x - 12)(x - 1)           (7x - 1)(x - 12)           (7x - 6)(x - 2)          (7x - 2)(x - 6)

     (7x - 4)(x - 3)             (7x - 3)(x - 4)

Check each of the above to find the proper factors.

If none of them work, then the problem cannot be factored using whole numbers.

We need to find a common denominator to figure this out. A common demoninator is where the bottom number of both fractions are the same.

You might be thinking, "5 isn't a fraction though!" Well, it turns out that any whole number can be written as a fraction if you put it over 1. So to write 5 as a fraction it would be:

$$\frac{5}{1}$$

So now we have:

$$\frac{13}{3} - \frac{5}{1}$$

There's a problem though, the bottom number isn't the same. To fix this, multiply the top and bottom of the second fraction by 3. You can do this as long as you do it to both the top and bottom.

$$\frac{13}{3} - \frac{5\times3}{1\times3}$$

$$\frac{13}{3} - \frac{15}{3}$$

Now that we have the common demoninator, we can subtract the tops of the fractions, and put that over 3.

$$\frac{13-15}{3}$$

$$\frac{-2}{3}$$

which is the same thing as:

$$\mathbf{-\frac{2}{3}}$$

A 1/2 exponent is a square root; 

so, 422 ^ (2)  means  √422  =  20.5426 ...

x is an unknown. We never know what x is until we solve. To do this we have to balance out both sides, so if we subtract 6 from one side, we have to subract 6 from the other side. If we multiply one side by 7, we have to multiply the other side by 7 too.

To solve this one, we have to add 2 to both sides to get x alone. You always want to get x alone because then you know what it equals.

23=x-2 

+2   +2

-------------

23+2=x

25=x

which means that

x = 25

3 because 1+!+!+11=1=1=+1 equals three

9x⁴ - 4y⁴  is the difference of two squares, so can be expressed as:

(3x² + 2y²)(3x² - 2y²)

Turn 6.9% into the fraction 6.9/100 and multiply this by 198:

6.9*198/100 = 13.662

2333333333333333333333333333333333333333333333

The angle the strings make with the horizontal must be 30°, so the radius of rotation must be

r = 1.59*cos(30°) metres

Resolve forces in vertical and horizontal directions.  These must each be balanced.

Vertical:  T1sin(30°) = T2sin(30°) + mg  ...(1)  

where T1 is tension in upper string (58.2N); T2 that in lower string; m is mass of ball (1.14kg) and g is gravitational acceleration (9.8m/s²)

Horizontal:  T1cos(30°)+T2cos(30°) = mv²/r ...(2) 

where v is speed of ball.

From (1)  T2 = T1 - mg/sin(30°)   ...(3)

                   =  58.2 - 1.14*9.8/sin(30°)

$${\mathtt{T2}} = {\mathtt{58.2}}{\mathtt{\,-\,}}{\frac{{\mathtt{1.14}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}} \Rightarrow {\mathtt{T2}} = {\mathtt{35.856}}$$

T2 ≈ 35.9 Newtons

Use (3) in (2)

T1cos(30°) + T1cos(30°) - mgcos(30°)/sin(30°) = mv²/r 

v² = (r/m)*(2T1cos(30°) - mgcos(30°)/sin(30°))

    = (1.59*cos(30°)/1.14)*(2*58.2*cos(30°) - 1.14*9.8*cos(30°)/sin(30°))

$${\mathtt{v}} = {\sqrt{\left({\frac{{\mathtt{1.59}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}}{{\mathtt{1.14}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{58.2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}{\mathtt{\,-\,}}{\frac{{\mathtt{1.14}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}}\right)}} \Rightarrow {\mathtt{v}} = {\mathtt{9.919\: \!048\: \!659\: \!810\: \!567}}$$

v ≈ 9.92 metres/sec

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1/8 as a decamal would be 0.125

The questions on the form I am not so great at math all the time 

Yes but I also watch vines and it is 21 jk it is 19 but I wanted to see what everyone would say

Let tension in the wire be T, mass be m, velocity be v, length of wire be L, radius be r and gravitational acceleration be g.

Resolve forces horizontally and vertically.

Vertical:  Tcos(θ) = mg  ...(1)

Horizontal:   Tsin(θ) = mv²/r    ...(2)

The radial acceleration is v²/r = Tsin(θ)/m = g*sin(θ)/cos(θ) = g*tan(θ) using (1) and (2)

Plug in the numbers!

The derivation is as follows.  You can plug in the numbers.

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Sat 4/10/14

1) BREAKING NEWS - THE TOWN CRIER IS ANNOUNCING A VERY IMPORTANT MESSAGE.

@@ End of Day Wrap       TFri 3/10/14        Sydney, Australia      Time 6:55pm      ♬  

Good evening everyone,

Today our fantastic answerers have been CPhill, Geno3141, alan, AzizHusain, angelicaMarrufo, Bugzy28871, Dragonslayer554, aoanewbie and The Ghost of Web3calc.   Thank you all.     

Aziz has visited us today.  That is great, we have all miss you Aziz.  

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acos is the same as inverse cos