dierdurst

Let's start with the price of the pants:

Both started at \($30\) but then were discounted by \(15%\)%. That means that the discounted price is \(85\)% of \(30\). Given this, we have \(30 \times 0.85 = 25.5\), which is our price. Now, we need to add a \(5\)% tax. We can get \(25.5 \times 1.05 = 26.775 ≈ 26.78\). Since it was asking for the total price, and Nona bought two pairs of pants, we would have \(26.78\times2 = $53.56\).

Next, let's move onto the discounts of a dress:

We have two cases;

Case 1: A \($50\) dollar dress that was discounted by \(20\)% and \(10\)%

Case 2: A \($50\) dollar dress that was discounted \(30\)%

Case 1: Both the coupons added would be \(0.90(0.80(50))\) which equals \(36\).

Case 2: We have \(0.70 \times 50\) which equals \(35\).

Wow, what a close call. The \(30\)% coupon saves \($1\) more than the \(20\)% and \(10\)% coupons. 

Finally, let's see how many possible outfits Nona can make:

Nona has \(9\) pairs of pants, \(12\) shirts and \(3\) belts, and an outfit can consist of a pair of pants, a shirt and a belt or no belt. 

We will start with the pants and shirts. In order to find all possible combinations, we can do \(9 \times 12 = 108\). We will save that number, and use it for our next step. In order to find the possible amount of outfits with a belt, we can do \(108 \times 3 = 324\). So \(324\) is the number of outfits WITH a belt, and \(108\) is the number of outfits WITHOUT a belt. When we add these, we get our answer, \(432\).

- Daisy

By the chart, you can see that there are:

Fifteen \(1\)'s

Fifteen \(2\)'s

Fifteen \(3\)'s

Fifteen \(4\)'s

Five \(5\)'s

Five \(6\)'s

Five \(7\)'s

Five \(8\)'s

Five \(9\)'s    and 

Four \(0\)'s (Of course, the zero's would not matter, because they are asking for the sum).

So we are trying to find \(s\), the sum of the digits. We can get the equation \(s = 15(1) + 15(2) + 15(3) + 15(4) + 5(5) + 5(6) + 5(7) + 5(8) + 5(9) \), which is equal to \(s = 15 + 30 + 45 + 60 + 25 + 30 + 35 + 40 + 45\), which we can finally get \(s = 325\)

- Daisy

For \(a23\), \(n = 22\). We can set up the equation \({2}^{22}\times a22 = 2p\). We can see that \(a22 = 2^{21}\times a21\). We can keep doing this until we get: \(2^{22+21+20...+3+2+1} \times 1 = 2^p\), so \(p = 22+21+20...+3+2+1 = 253\). 

- Daisy

Please make sure you have posted the whole question, some of your question is cut off. 

- Daisy

Yeah, they are fun!

- Daisy

Since there is no common base number, we will have to make a common exponent number. The GCF of all of these exponents is \(100\), so lets make all of the exponents \(100\). We have:

a) \({25}^{100}\)

b) \({2}^{3(100)} = {8}^{100}\)

c) \({3}^{4(100)} = {81}^{100}\)

d) \({4}^{2(100)} = {16}^{100}\)

e) \({2}^{6(100)} = {64}^{100}\)

By looking at the base numbers, we can tell which is the least and which is the greatest. The numbers from greatest to least is: \(3^{400} , 2^{600}, 25^{100} , 4^{200} , 2^{300}\)

- Daisy

We can make \(x^2 - y^2 = -12\) into \((x-y)(x+y) = -12\) by using “Difference of Squares”. Since we know that \(x+y = 6\), we can substitute that in. Now we have \(6(x-y) = -12\). We can divide \(6\) on both sides, which gives us \(x - y = -2\). We also still have the equation \(x + y = 6\). We can use elimination to get rid of \(x\). We will subtract the second equation from the first. This gives us \(-2y = -8\), so \(y = 4\). We can plug this back into any equation to get \(x\). So \(y = 4\) and \(x = 2\).

- Daisy

That's great! If you still have any questions, feel free to still ask.

- Daisy

No problem! 

- Daisy