Answer:
Step-by-step explanation:
5
=15.70...
To round a number to the nearest tenth , look at the next place value to the right (the hundredths)which is 0 in this question.
Thus, it will become 15.7
18.75÷5=3.75
12.75÷3=4.2
The pack of 5 books has the lower cost per book
Try this option:
1. Common view of hyperbola equation is:
![\frac{x^2}{a} - \frac{y^2}{b} =1 \ , where \ a,b-numbers](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%5E2%7D%7Ba%7D%20-%20%5Cfrac%7By%5E2%7D%7Bb%7D%20%3D1%20%5C%20%2C%20where%20%20%5C%20a%2Cb-numbers)
2. hyperbolas are: 2x²+4x-5y²-10y+57=0 and -x²+12x+3y²+7y+11=0.
Answer:
A bit of a doosy, but bare with me: f*g*h(x) = ![x^2+54x-12x\sqrt{x} -108\sqrt{x} +90](https://tex.z-dn.net/?f=x%5E2%2B54x-12x%5Csqrt%7Bx%7D%20-108%5Csqrt%7Bx%7D%20%2B90)
Step-by-step explanation:
So explaining this is a bit awkward as well, but I can try...So
Starting from g(x) working that backwards into f(x), so...
g(x) = ![x-3](https://tex.z-dn.net/?f=x-3)
f(x) = ![(x^4+9)](https://tex.z-dn.net/?f=%28x%5E4%2B9%29)
Take your g(x) and plug it into your f(x) into every x you see
f(g(x)) =
~ I hope you can visualize this, but do you see how g(x) plugs into f(x)...
Now you still have one more step which you still need to include h(x), so
Remember h(x) = ![\sqrt{x}](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D)
Now plug in your h(x) into every x you see from f(g(x)) and it should look like...
![(\sqrt{x} -3)^4 + 9](https://tex.z-dn.net/?f=%28%5Csqrt%7Bx%7D%20-3%29%5E4%20%2B%209)
> is the symbol for greater than