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It's a 

because a Quadratic equation uses squares (^2) and a Cubic equation uses cubes (^3)

Well then I'm sorry I can't help you. I really don't understand how the answer is 40 unless I understand the question incorrectly 

That is the full question.

Can you type the full question? Because maybe I'll understand it better if I get the full question

No that's why the question says

did you type in the full question? I think were some parts you didnt add.

Yea the answer is 40 but I don't know how to get 40

I'm sorry. I don't understand. Can you please clarify what you mean by 40? Did I understand the question wrong?

How to get 40?

We can make 2 equations and solve for the variables.

Lets make l = length and w = width.

\(w=1/4*l\)  

\(2w+2l=80\)

You can plug in w = 1/4*l into the second equation, so it turns into \(2(1/4l)+2l=80\)

Solving for that equation, you get \(l=32\)

So your answer would be 

The question isn't very clear but I'll answer based on my understanding. So Hunter spent 1/4 of his money and an additional 7 dollars for something else. He has 23 dollars left. And you want to find how much he had in the beginning. If this is the question, the answer's below.

You should make an equation with the variable x for how much he had at first. So it would be \(1/4x + 7 = 23\)

Solving for the equation, you get 

You have to move \(8x^4 - 7x^2 + 8x - 7 \) to the right side so it becomes \(g(x) = (x^6 - x^3 - 2x^2 + 1) - (8x^4 - 7x^2 + 8x - 7)\)

Simplify

Therefore, your answer would be

The probability is 5/16.

The real numbers that are not in the domain of the function f(x)=(1x−64)+(1x2−64)+(1x3−64)f(x)=(1x−64)+(1x2−64)+(1x3−64) are the numbers that will make the function undefined.

We look at the fraction denominators. The function will be undefined when either of the denominators is equal to zero.

Case 1:

x−64=0x−64=0

Add 64 to both sides and simplify:

x−64+64=0+64⟹x=64x−64+64=0+64⟹x=64

Case 2:

x2−64=0x2−64+64=0+64x2=64⟹x=±√64∴x=8,x=−8x2−64=0x2−64+64=0+64x2=64⟹x=±64∴x=8,x=−8

Case 3:

x3−64=0x3−64+64=0+64x3=64⟹x=3√64∴x=4x3−64=0x3−64+64=0+64x3=64⟹x=643∴x=4

Hence the real numbers that are not in the domain of the given function are −8,4,8−8,4,8 and 64

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This is a pretty long question, but I think I'm up for answering it.

First, we need to multiply the two expressions given, and then combine like terms.

(24x^4 + 6x^3 + 4x^2 + 7x - 5) * (6x^3 + 3x^2 - 3x + 4)

I'm going to switch them around to make it easier:

(6x^3 + 3x^2 - 3x + 4) * (24x^4 + 6x^3 + 4x^2 + 7x - 5)

= 6x^3 (24x^4 + 6x^3 + 4x^2 + 7x - 5) + 3x^2 (24x^4 + 6x^3 + 4x^2 + 7x - 5) - 3x (24x^4 + 6x^3 + 4x^2 + 7x - 5) + 4 (24x^4 + 6x^3 + 4x^2 + 7x - 5)

= (144x^7 + 36x^6 + 24x^5 + 42x^4 - 30x^3) + (72x^6 + 18x^5 + 12x^4 + 21x^3 − 15x^2) + (−72x^5 − 18x^4 − 12x^3 − 21x^2 + 15x) + (96x^4 + 24x^3 + 16x^2 + 28x − 20)

= 144x^7 + 36x^6 + 24x^5 + 42x^4 - 30x^3 + 72x^6 + 18x^5 + 12x^4 + 21x^3 − 15x^2 −72x^5 − 18x^4 − 12x^3 − 21x^2 + 15x + 96x^4 + 24x^3 + 16x^2 + 28x − 20

= 144x^7 + 36x^6 + 72x^6 + 24x^5 + 18x^5 − 72x^5 + 42x^4 + 12x^4 − 18x^4 + 96x^4 - 30x^3 + 21x^3 − 12x^3 + 24x^3 − 15x^2 − 21x^2 + 16x^2 + 15x + 28x − 20

= 144x^7 + 108x^6 - 30x^5 + 132x^4 + 3x^3 - 20x^2 + 43x - 20

So, now let us figure out the coefficient of x^3. We can see in our final answer that we have 3x^3, so the coefficient would be 3.

Therefore, our answer is 3.

P.S. I certainly hope this helps, but as it was an extremely time-taking and quite difficult problem, my answer may be incorrect. I do believe I worked through it correctly, though, so I hope this helps :)

your answer is 

First a is chosen at random from set {1,2,3,...,99,100}, and then b is chosen at random from the same set. What is the probability that the integer 3^a+7^b has units digit 8?

I looked this up quite a bit, and most people seem to think the answer is "odd numbers".

Let me explain:

Because the number of months in a year are 12, we can take the two numbers in 12 (1 and 2).

(1), 2 = 1 odd number

So, once in a year.

In a month, there are four weeks. So we take the four numbers leading up to four (1, 2, 3, and 4).

(1), 2, (3), 4 = 2 odd numbers.

So, twice in a month.

In a week, there are seven days. So we take the seven numbers leading up to seven (1, 2, 3, 4, 5, 6, and 7).

(1), 2, (3), 4, (5), 6, (7) = 4 odd numbers.

So, four times in a week.

Finally, in a day, the people who have answered this question before seem to assume that there are 12 hours during daytime. We can take all 12 numbers leading up to 12 (1, 2, 3, 4 ... 12)

(1), 2, (3), 4, (5), 6, (7), 8, (9), 10, (11), 12 = 6 odd numbers

So, six times in a day.

I'm not entirely sure that this answer is correct, because there are a few assumptions we had to make, but this is the best answer I could find.

Hope this helps :)

Your answer makes no sense to me Hannah.  (MG)

Start by using the identity

sin(A-B)=sinAcosB-sinBcosA

https://books.google.com/books?id=ks10DwAAQBAJ&pg=PA323&lpg=PA323&dq=Troy+was+a+member+of+the+DVD+store+last+year.+Last+year,+he+spent+52.50+zeds+in+total,+which+included+his+membership+fee.+How+much+would+Troy+have+spent+if+he+had+not+been+a+member+but+had+rented+the+same+amount+of+DVDs?&source=bl&ots=cTZTMfps9W&sig=ACfU3U1uoms67-6Ww-i7r0I2h3Lq9ICXHg&hl=en&sa=X&ved=2ahUKEwiEudKIxZ7wAhXJITQIHSgUDQEQ6AEwAXoECAIQAw#v=onepage&q=Troy%20was%20a%20member%20of%20the%20DVD%20store%20last%20year.%20Last%20year%2C%20he%20spent%2052.50%20zeds%20in%20total%2C%20which%20included%20his%20membership%20fee.%20How%20much%20would%20Troy%20have%20spent%20if%20he%20had%20not%20been%20a%20member%20but%20had%20rented%20the%20same%20amount%20of%20DVDs%3F&f=false

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The answer is the letter 'E'. Once in the word yEar. Six times in “EvEry wEEkEndEr. (from google)

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The answer is E

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1000 1/3 is alr simplified, but you can make it a improper fraction. 

1000 1/3 = (1000*3+1)/3 = 3001/3

=^._.^=

2&3) you need to go through these one by one and draw out the function. Like x^4 is a parabola, so it falls into the third category as it is not flat or increasing all the time, and it's not flat or decreasing all the time - first it goes down, then it goes up.

X^ 4 is also an even function, because it gives the same value whether x is positive or negative (it's symmetrical over the y axis). Something like x^3 would be odd since (-x)^3 = -(x^3). If you draw a diagonal line through the origin on your graphs it's symmetrical over that line. Hope this helps

6 of 5.98 :))

5.98 * 6 

=^._.^=