All Answers 358090 Answers

Emily, i think we will solve them like 

1). 2x(x+3)+4(x-1)  So now we will multiply 2x by x and 3 which gives us 2x^2 and 6x!And also multiply 4 by

x and -1 which gives us 4x and -1x  

=$$2x^2 + 6x+4x-4$$  ( so now we will simply multiply or subtract them as they are placed)(here we just have to add 6x and 4x )

$$=2x^2+10x-4$$ $$ANSWER$$

I hope u've understood by now but if not then pls be free to ask!

hope this helps!

Im working on the other ones and post my answer in this same post when i solve them!

$$\begin{array}{rlll}
[sec^6(x)]-[tan^6(x)]&=&1+[3tan^2(x)]+[3tan^4(x)]\\\\
(\frac{1}{cos^6x})-[(\frac{sin^6x}{cos^6x})]&=&1+[3(\frac{sin^2x}{cos^2x})]+[3(\frac{sin4^x}{cos^4x})]\\\\
cos^6(x)((\frac{1}{cos^6x})-[(\frac{sin^6x}{cos^6x})])&=&cos^6(x)(1+[3(\frac{sin^2x}{cos^2x})]+[3(\frac{sin^4x}{cos^4x})])\\\\
(1)-[sin^6x]&=&cos^6(x)+[3(sin^2xcos^4(x))]+[3(sin^4xcos^2(x))]\\\\
1-[(sin^2x)^3]&=&cos^6(x)+[3(sin^2x)cos^4(x)]+[3((sin^2x)^2)cos^2(x)]\\\\
1-[(1-cos^2x)^3]&=&cos^6(x)+[3(1-cos^2x)cos^4(x)]+[3((1-cos^2x)^2)cos^2(x)]\\\\
1-[(-cos^2x)^3+3(-cos^2x)^2+3(-cos^2x)+1]&=&cos^6(x)+[3(1-cos^2x)cos^4(x)]+[3((cos^2x)^2-2cos^2x+1)cos^2(x)]\\\\
1-[(-cos^6x)+3(cos^4x)-3cos^2x+1]&=&cos^6(x)+[3(cos^4(x)-cos^6(x))]+[3((cos^6x)-2cos^4x+cos^2(x))]\\\\
1+cos^6x-3cos^4x+3cos^2x-1&=&cos^6(x)+[3cos^4(x)-3cos^6(x)]+[3cos^6x-6cos^4x+3cos^2(x)]\\\\
cos^6x-3cos^4x+3cos^2x &=&cos^6(x)+3cos^4(x)-3cos^6(x)+3cos^6x-6cos^4x+3cos^2(x)\\\\
-3cos^4x &=&-3cos^4(x)-6cos^4x\\\\
-3cos^4x &=&-3cos^4x\\\\
LHS&=&RHS\\\\
\end{array}$$

$$\mbox{This is true for all values of x}$$

Emily , these are a lot!And if we do all of them for you then how will you learn!So we cant do all of them so why dont you number them and then tell us the important ones which we can do for you so that you understand!!

Answer deleted!

5+5=10

and you could also use the on-site calculator for this aswell!

Thank you AzizHusain!

im happy you enjoyed my post!!!

I just did a HUGE page of LaTex on this and lost the whole blink'n lot.

Bummer!!!!!!

ok When I did it on paper I got down to cosx=0

When I did it straight onto the forum I got that it was always true x=anything

I'll redo it a little later.  This is hugely irritating.

Anyway the way I went about it was to change everything to sin and cos

then mult by cos^6x

Then get rid of all the sines

The collect like terms 

and wallah! it will all fall out.

sec^6(x)-tan^6(x)-3*tan^4(x)-3*tan^2(x)-1=0

No problem and yes, they do change results dramatically sometimes.

AzizHusain, thank you, your notation was correct. The answer was approximately 46.49.  Funny how a minor thing like paranthesese can change the entire result.

I believe you need to upload the image, but to do that, you need to become a member first by registering first on the website before you can upload attachments/files when you post questions.

Perform the FOIL method:

First - 2*4sqrt(5) = 8sqrt(5)

Outside - 2*(-2) = -4

Inside - 4sqrt(10)*4sqrt(5) = 16sqrt(50)

Last - 4sqrt(10)*(-2) = -8sqrt(10)

Adding all the terms: 8sqrt(5) - 4 + 16sqrt(50) - 8sqrt(10)

16sqrt(50) can be converted to: 16*sqrt(25*2) = 16*sqrt(25)*sqrt(2); we can take out the 25 from the square root as a 5 since sqrt(25) = 5 --> 16*5sqrt(2) = 80sqrt(2)

Now, we have 8sqrt(5) - 4 + 80sqrt(2) - 8sqrt(10)

8[sqrt(5) - sqrt(10) + 10sqrt(2) - (1/2)]*

*I have just done the distributive property backwards so that when you distribute it, you get the same result: 8sqrt(5) - 4 + 80sqrt(2) - 8sqrt(10)

For 80sqrt(2), since 8 is a common term that can be taken out, we can make it 8(10sqrt(2)).

For -4, we know that 8*(-1/2) = -4; therefore, we can make it 8(-1/2).

Now, combining like terms for the 8sqrt(5) and - 8 sqrt(10) is easy since both have an 8. 

That is how we end up with 8[sqrt(5) - sqrt(10) + 10sqrt(2) - (1/2)] as the final answer.

E = 1/2mv^2

First, multiply both sides by 2 to get rid of the 1/2 on the right-hand side of the equation: 2E = mv^2

Next, divide by v^2 to obtain: 2E/v^2 = m.

A square foot is a unit of area and simply expressed as ft^2 for simplification. 1 sq ft or ft^2 would be a square that is 1 foot by 1 foot since 1*1 = 1 ft^2. Each "side" or length & width would be 1 feet.

We begin by squaring both sides to get rid of the square roots: sqrt(6x - 15) = sqrt(4x + 7) --> 6x - 15 = 4x + 7

Next, we get all the x terms on one side and all the regular numbers to the other side by adding 15 to both sides and subtracting 4x from both sides: By doing this, we get 2x = 22.

Now, we are left with something easy and we can see that 22/11 = x --> x = 11.

Do a quick check to see if x = 11 satisfies the condition:

sqrt(6*(11) - 15) = sqrt(4*(11) + 7) 

sqrt(66 - 15) = sqrt(44 + 7)

sqrt(51) = sqrt(51) --> Yes.

We can use proportions to solve this problem. 

Proportions are useful since they can be used to relate units and any missing pieces can be found.

In this case, we need to have this format: _ KG / _ N = _ KG / _ N

So far, we have this much information: 7 KG / 35 N = 8 KG / _ N

If we cross multiply, we obtain: (7 KG* _ N) = (8 KG*35 N) --> (7KG* _ N) = (280 KGN)

Dividing by 7 KG, we cancel the KG's out and obtain: 40 N.

Note: If 7 kg = 35 N, 8 kg = 40 N, 9 kg = 45, etc. We can see a pattern (For every 1 KG increase, we see an increase of 5 N). Furthermore, you could just multiply the mass by 5 to obtain the N, the formula would be z(KG) = z*5(N) where z = the mass. So for 90 KG, we would have 90*5 = 450 N.

Interesting nicknames and fun post. It is nice to relax at the end of the day or whenever and enjoy something new. I have never been on such an active forum before. Also, you posted some cool pictures which were all incredibly funny and awesome! 

It's possible that log(5)x is meant to be log to the base 5 of x.  If so, and we are talking real numbers only then the following graph shows the x-intercept.

Yes, easy to overlook this, but it's always worth keeping an eye out for the difference of two squares, even when there isn't an explicit ² sign on the number 1. 

Ohh, I see the simplification part now! Thanks, Alan! I overlooked splitting the numerator into two.

$$\frac{\tan^2a - 1}{1+\tan a }=\frac{(\tan a + 1)(\tan a -1)}{1+\tan a}=\tan a -1$$

sqrt(-49) = 7i

sqrt(-64) = 8i

We have imaginary numbers since square roots of negative numbers cannot be real numbers. We use the "i" notation instead.

7i*8i = 56(i^2)*

*The multiplication of i by i = i*i = i^2 = -1

We get 56(-1) = -56.

Represent the cost of an adult ticket by 'at' and the cost of a student ticket by'st'.

For the first family    2*at + 1*st  =  8.75

For the 2nd family    1*at + 4*st  = 10.50

From the 2nd family's equation, if we subtract 4*st from both sides we have 1*at = 10.50 - 4*st, which tells us the cost of an adult ticket in terms of the cost of student tickets.  The first family paid for 2 adult tickets which means they must have paid the equivalent of 2*(10.50 - 4*st) or 21 - 8*st for their 2 adult tickets.  Replacing the 2*at in their equation we get, for the first family, that:

 21 - 8*st + 1*st = 8.75

or 21 - 7*st = 8.75

Add 7*st to both sides and subtract 8.75 from both sides;

21 - 8.75 = 7*st

12.25 = 7*st

Divide both sides by 7 to find

st = 12.25/7 =  1.75

So the price of a student ticket is $1.75

Put this back in the equation for the 2nd family and we get

1*at + 4*1.75 = 10.50

at + 7 = 10.50

Subtract 7 from both sides

at = 10.50 - 7 = 3.50

so the price of an adult ticket is $3.50