Answer:
Answer:
18 cups.
Step-by-step explanation:
We are given that Bernie spends $6.50 on ingredients and cups for his lemonade stand. He charges $1.50 for each cup of lemonade. Inequality that represents this situation: .
To find number of cups x to make a profit of at least $20 we will use our given inequality.
Therefore, in order to make a profit of at least $20 Bernie need to sell 18 cups of lemonade.
Answer:
No. The new height of the water is less than the height of the glass(6.33 cm<10 cm)
Step-by-step explanation:
-For the water in the glass to overflow, the volume of the inserted solid must be greater than the volume of the empty space or the ensuing height of water >height of glass.
#Volume of the golf ball:
![V=\frac{4}{3}\pi r^3\\\\=\frac{4}{3}\pi \times 4^3\\\\\approx 268.08\ cm^3](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3%5C%5C%5C%5C%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%5Ctimes%204%5E3%5C%5C%5C%5C%5Capprox%20268.08%5C%20cm%5E3)
#The volume of the water in the glass:
![V=\pi r^2 h\\\\=\pi \times 4^2\times 10\\\\\approx 50.27\ cm^2](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2%20h%5C%5C%5C%5C%3D%5Cpi%20%5Ctimes%204%5E2%5Ctimes%2010%5C%5C%5C%5C%5Capprox%2050.27%5C%20cm%5E2)
We then equate the two volumes to the glass' volume to determine the new height of the water:
![V=\pi r^2h\\\\(206.08+50.27)=\pi r^2 h\\\\h=318.35/(\pi \times 4^2)\\\\=6.33\ cm](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2h%5C%5C%5C%5C%28206.08%2B50.27%29%3D%5Cpi%20r%5E2%20h%5C%5C%5C%5Ch%3D318.35%2F%28%5Cpi%20%5Ctimes%204%5E2%29%5C%5C%5C%5C%3D6.33%5C%20cm)
Hence, the glass will not overflow since the new height of the water is less than the height of the glass(6.33 cm<10cm).
Step-by-step explanation:
∫₀² x f(x²) dx
If u = x², then du = 2x dx, and ½ du = x dx.
When x = 0, u = 0. When x = 2, u = 4.
∫₀⁴ ½ f(u) du
½ (16)
8
Answer:
x≠-1,1
Step-by-step explanation:
f(g(x)) is a composition where g(x) is is substituted for x in f(x).
Recall, f(x) is 2/x. So we write 2/|x|-1. This places x in the denominator and 0 cannot be in the denominator x. Any value of x that makes the denominator 0 will not be in the domain.
|x|-1=0
|x|=1
x=1,-1