dierdurst

The profit of the screwdriver is \($2.00\), and we need to sell enough screwdrivers to make a profit of \($124.00\). We can solve this by doing \(\frac{124}{2} = 62\). So you need to have \(62\) screwdrivers produced.

- Daisy

For the first two, you can use Pythagorean Theorem. This states that \({a}^{2}+{b}^{2}={c}^{2}\), where \(a\) and \(b\) are the legs, and \(c\) is the hypotenuse. 

Problem 1: We already have sides \(a = 10\) and \(c = \sqrt{170}\). We can move the formula around for it to fit this problem better. We can change it to \({b}^{2}={c}^{2}-{a}^{2}\). When subsituting our values in, we can get \({b}^{2}={\sqrt{170}}^{2}-{10}^{2}\). When we square both values, we can get \({b}^{2} = 170-100\), so \({b}^{2} = 70\). We can solve this to be \(b = \sqrt{70}\) or \(x = \sqrt{70}\).

Problem 2: We already have sides \(b = 5\sqrt{2}\) and \(c = 13\). We can move the formula around for it to fit this problem better. We can change it to \({a}^{2}={c}^{2}-{b}^{2}\). When plugging in our values, we can get \({a}^{2} = {13}^{2} - ({5\sqrt{2})}^{2}\). By solving this, we can get \({a}^{2} = 119\) which equals \(a = \sqrt{119}\) or \(x = \sqrt{119}\).

- Daisy

From these two sentences, we can get two equations: \(3s + 7t = 134\) and \(3s + 4t = 92\). In order to solve this, we can use elimination, to get rid of one of the variables. If we subtract the second equation from the first, we can get \(3t = 42\), which simplifies to \(t = 14\). Now that we know that the student tickets cost $14, we can plug that into either the first or second equation. Both will give the same answer. Let's just use the first equation. From here we can get \(3s + 7(14) = 134\), which we can solve and get \(s = 12\). So the senior tickets cost $12.

- Daisy

Haha I'm fine, thank you very much, though! I'll keep that in mind the next time I have trouble on a question.

- Daisy

It's well drawn, and it helped me comprehend the solution. Thanks!

- Daisy

Thanks! That answered my question about the diagram! Sorry for taking up your time!

- Daisy

Problem \(1\):

The formula for the surface area of a cone is \(πrl+π{r}^{2}\), where \(r\) is the radius and \(l\) is the slant height. For problem one, \(r = 6\) and \(l = 14\). By plugging in the variable values, we can get \(π(6)(14)+π{6}^{2} = 84π + 36π = 120π\). 

Problem 2:

The formula for the surface area of a cylinder is \(2πrh+2π{r}^{2}\), where \(r\) is the radius and \(h\) is the height. \(r = 13\) and \(h = 17\). When you plug in the variable values, you can get \(2π(13)(17)+2π{13}^{2} = 442π + 338π = 780π\), and it is asking for the nearest whole number, so multiply \(3.14\) by \(780\). The product is \(2449.2\), which rounds down to \(2449\).

Problem 3:

The formula for the surface area of a cone is \(πrl+π{r}^{2}\), where \(r\) is the radius and \(l\) is the slant height. \(r = 7\) and \(l = 24\). By plugging in the variable values, we can get \(π(7)(24)+π{7}^{2} = 168π + 49π = 217π\). Since they are asking for a decimal to the nearest tenth, we can get \(681.4\).

Problem 4:

The formula for the surface area of a cylinder is \(2πrh+2π{r}^{2}\), where \(r\) is the radius and \(h\) is the height.  \(r = 25\) and \(h = 36\). When you plug in the variable values, you can get \(2π(25)(36)+2π{25}^{2} = 1800π + 1250π = 3050π\).

- Daisy

I have figured it out! Thank you so much for your help.

- Daisy

Haha! Thanks for informing me, Melody! I will keep that in mind in the future.

- Daisy

In order to solve an equation like this, we need to isolate the variable. To do this, we need to get rid of \(5\) from \(x\). Since \(5\) is being multiplied to \(x\), we need to do the opposite operation, which is division. When we divide \(5\) from \(x\), we also have to divide \(5\) from \(25\). When \(5\) divides \(25\)​, we get \(5\), so your result is \(x = 5\).

- Daisy