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It doesn't say that there's a multiplication sign in the middle of 5 and (7-2). Adding a sign there would make the equation different. It would be regrouped into:

60 ÷ 5 × ( 7 - 5 ) = ( 60 ÷ 5 ) ( 7 - 5 )

Which is different from the original equation.

But if no multiplication sign was added, it would be purely '60 divided by the product of 5 and 7 minus 2' which is 6.

60 ÷ 5 ( 7 - 2 )

60 ÷ 5 ( 2 )

60 ÷ 10

6

What you're saying here is that the equation is 60 over 5 THEN multiply by 2 instead of 60 over 5(2)

If you legit put the equation into a real life TI-84 calculator as "60/5(7-5)" you'll get 6. The scientific calculator on the computer is incompatible with that equation because it forces you to place the multiplication symbol in between the 5 and the (.

Nor should they have believed you!  See my reply #5.

No Gh0sty!  

Parentheses first:  60/5*2

Divide and Multiply have the same precedence so you work from left to right:

60/5*2 → 12*2 → 24

Yeah. I kept arguing they have to deal with the Parantheses first, PEMDAS, and all, but their logic got them to 60/5*(2) so then they followed that with 12*2 = 24. I kept telling them there was no "*" sign so they had to multiply first because they still have to deal with parantheses first, but alas, they didn't believe me.

The domain refers to all allowable values of x.  Here that is from -infinity to + infinity (R)

The range refers to the resulting values of y (or fog(x)).  Since (x^2 -1)/(x^2 + 1) can range from -1 (when x is 0) to +1 (when the absolute value of x is very large) then the angle whose cosine is (x^2 -1)/(x^2 + 1) can be any positive or negative value. So the range is -infinity to +infinity as well (R).

this is following answers :

1) [0,1]

2) [-1/2,1/2]

3) R

4) [-1,1]

Did you say it was 6?

THANK YOU. My friends have been arguing with me about this for eons.

tnx but i can't find domain 

So we have the equation 60/5(7-5).

First simplify the '7-5' in the brackets to get 60/5(2)

Simplify again and get 60/10

And that's just dividing 60 by 10, so the answer is 6.

6 is our answer.

Not 24.

New update: I think this answer is wrong. Look at Alan's answer. Sorry. 

"if f(x)=cos^-1(2x-1) and g(x)=x^2/1+x^2 find domain of fog(x)??"

The following should help:

.

Hello 

Assume that the student has a cup with 12 writing implements: 6 pencils, 4 ball point pens, and 2 felt-tip pens. In how many ways can the selection be made if no more than one ball point pen is selected if they must choose 4 implements?

\(\begin{array}{|rcll|} \hline && \binom{4}{1}\binom{8}{3} + \binom{4}{0}\binom{8}{4} \\ &=& 4\binom{8}{3} + 1\binom{8}{4} \\ &=& 4\cdot \frac{8}{3}\cdot \frac{7}{2}\cdot \frac{6}{1} + \frac{8}{4}\cdot \frac{7}{3}\cdot \frac{6}{2}\cdot \frac{5}{1} \\ &=& 4\cdot 8 \cdot 7 + 7 \cdot 2 \cdot 5 \\ &=& 224 + 70 \\ &=& 294 \\ \hline \end{array}\)

Help! I don't know how to solve these 2 questions. (Algebra II)

3.

\(\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 5} && 4 && -1 && -10 && -23 && -40 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = -1} && -5 && -9 && -13 && -17 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = -4} && -4 && -4 && -4 && \cdots \\ \end{array} }\)

\(\begin{array}{rcl} y &=& \binom{x}{0}\cdot {\color{red}d_0 } + \binom{x}{1}\cdot {\color{red}d_1 } + \binom{x}{2}\cdot {\color{red}d_2 } \\ \end{array}\)

\( \begin{array}{|rcl|} \hline y &=& \binom{x}{0}\cdot {\color{red} 5 } + \binom{x}{1}\cdot {\color{red} (-1) } + \binom{x}{2}\cdot {\color{red} (-4) } \\ &=& 5 - x - \frac{x}{2}\cdot \frac{x-1}{1} \cdot 4 \\ &=& 5 - x - 2x(x-1) \\ &=& 5 - x - 2x^2+2x \\ \mathbf{y} & \mathbf{=} & \mathbf{5 + x - 2x^2} \\ \hline \end{array}\)

4.

\(\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 20} && 4 && 0 && 20 && 76 && 180 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = -16} && -4 && 20 && 56 && 104 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 12} && 24 && 36 && 48 && \cdots \\ \text{2. Difference } &&&& {\color{red}d_3 = 12} && 12 && 12 && \cdots \\ \end{array} }\)

\(\begin{array}{rcl} y &=& \binom{x}{0}\cdot {\color{red}d_0 } + \binom{x}{1}\cdot {\color{red}d_1 } + \binom{x}{2}\cdot {\color{red}d_2 } + \binom{x}{3}\cdot {\color{red}d_3 } \\ \end{array}\)

\(\begin{array}{|rcl|} \hline y &=& \binom{x}{0}\cdot {\color{red} 20 } + \binom{x}{1}\cdot {\color{red} (-16) } + \binom{x}{2}\cdot {\color{red} 12 } + \binom{x}{3}\cdot {\color{red} 12 } \\ &=& 20 - 16x + \frac{x}{2}\cdot \frac{x-1}{1} \cdot 12 + \frac{x}{3}\cdot \frac{x-1}{2}\cdot \frac{x-2}{1} \cdot 12 \\ &=& 20 - 16x + 6x(x-1) + 2x(x-1)(x-2) \\ &=& 20 - 16x + 6x^2-6x + 2x \left(x^2-3x+2 \right) \\ &=& 20 - 16x + 6x^2-6x + 2x^3-6x^2+4x \\ &=& 20 - 16x -6x + 2x^3 +4x \\ \mathbf{y} & \mathbf{=} & \mathbf{20 -18x +2x^3} \\ \hline \end{array}\)

Use differences to find a pattern in the sequence
4,4,9,20,38,64,99, ...
Assuming that the pattern? continues,
the eighth term should be?

\(\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & 4 && 4 && 9 && 20 && 38 && 64 && 99 && {\color{red} 144} && \cdots \\ \text{1. Difference } && 0 && 5 && 11 && 18 && 26 && 35 && {\color{red} 45} && \cdots \\ \text{2. Difference } &&& 5 && 6 && 7 && 8 && 9 && {\color{red} 10} && \cdots \\ \text{3. Difference } &&&& 1 && 1 && 1 && 1 && {\color{red} 1} && \cdots \\ \end{array} } \)

\(\begin{array}{|rcrcr|} \hline 9 &+& {\color{red}1} &=& {\color{red}10} \\ 35&+&{\color{red}10} &=& {\color{red}45} \\ 99&+&{\color{red}45} &=& {\color{red}144} \\ \hline \end{array} \)

The eighth term is 99 + 35 + 9 + 1 = 144

rate  =  area painted / number of hours

rate  =  15.2 m²  /  4.6 hours

                                                      Divide the numerator and denominator by  4.6 .

rate  ≈  3.304 m² / 1 hour

rate  ≈  3.304 m² per hour

Is this the question?

    \(\frac6{24}\div\frac33\)

=  \(\frac{6\div3}{24\div3}\)

=  \(\frac28\)

I don't know if you want to do this, but we can further reduce this fraction by  2  , to get...

 =  \(\frac14\)

1)   Use differences to find a pattern in the sequence. 4,4,9,20,38,64,99 Assuming that the pattern continues, the eighth term should be?

If you look at the difference between the terms, you get the following:

0, 5, 11, 18, 26, 35......

Now, see the difference between these new terms and you get:

5, 6, 7, 8, 9.......etc. It, therefore, follows that you will add: 10, 11, 12, 13......etc. to the FIRST differences: 35+10 =45,  45+11=56, 56+12=68.........etc. Therefore, your sequence will be:

4,4,9,20,38,64,99, 144, 200,  268........etc.

P.S. You could use this more complicated formula to generate each term:

a(n) = 1/6 (n^3 + 9n^2 - 34n + 48).

2)  Use differences to find a pattern in the sequence. 6,9,20,39,66,101,144 Assuming that the pattern continues, the eighth term should be?

Do the same for this one and you will see these differences:

3, 11, 19, 27, 35, 43..............

Now look at the differences between these and you get a CONSTANT of 8. So, you will have:

43+8=51,  51+8=59, 59+8=67.........etc. So your sequence will be:

 6,9,20,39,66,101,144, 195, 254, 321.........etc. 

P.S. You could also use this more complicated formula to generate each term:

a(n) = 4n^2 - 9n + 11 

The backyard looks something like this.....

The area of the lawn is just the area of the backyard minus the area of the pond .

area of lawn   =                    area of backyard                - area of pond

area of lawn   =   (backyard length) * (backyard width)  -  area of pond

Plug in the information given in the problem.

area of lawn   =   25 m  *  10 m  -  27 m²

area of lawn   =   250 m²  -  27 m²

area of lawn   =   223 m²          

Use the formula S​ =n2 to find the sum of 1​ + 3​ + 5​ + ...​ + 597 = 

​(Hint: To find​ n, add 1 to the last term and divide by​ 2.)

n = [1 + 597] / 2 =299 terms. Or, [597 - 1] / 2 + 1 = 299 terms

Sum =[First + Last] / 2 x n

Sum = [ 1 + 597] / 2 x 299

Sum =       299 x 299

Sum = 89,401

Use the formula S​ = n2 to find the sum of 1 + 3 + 5 + ... + 103 = (Hint: To find​ n, add 1 to the last term and divide by​ 2.)?

n = [1 + 103] /2 = 52 terms. Or, 103 - 1 = 102/2 + 1 = 52 terms.

Sum = [First + Last] / 2 x n

Sum =[ 1 + 103] / 2 x 52

Sum =     52 x 52

Sum =2,704 

Hmmm... I'm sorry for not explaining the exponent towers earlier. I'll try my best to give a logical explanation on why exponent towers should be evaluated from the top downward despite it possibly being counterintuitive. To answer your original question, \(10^{10^{100}}=10^{\left(10^{100}\right)}\)

Let's use the variables a, b, and c and suppose that \(a^{b^{c}}=\left(a^{b}\right)^c=a^{bc}\), then the existence of exponent towers would be completely unnecessary as only one level of the exponent would ever be necessary; there must be a reason for such notation to exist. 

Another reason is that it is consistent with the order of operations. Exponents is a level below parentheses when it comes to priority, but the order of operations applys to the power, too. Here's an example:

\(2^{2+3}=2^5=32\)

\(3^{2*2}=3^4=81\)

And therefore, you must evaluate the exponent within the exponent. Does this make sense?

I learned today that even modern-day programs calculate have ambiguity in the matter. Excel 2013 is an example, as \(2^{3^2}\) outputs 64 as opposed to 256. In the future, 

Don't have to solve this I already got the answer!