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Ivahew [28]
3 years ago
15

3. How many rides can be taken with the $100 pass?

Mathematics
1 answer:
Feliz [49]3 years ago
4 0

Answer:

89.7898

Step-by-step explanation:

don't use this awnser its wrong

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a cone has a volume of 6 cubic inches. what is the volume of a cylinder that the cone fits exactly inside of
ella [17]

Answer:

= 18 cubic inches

Step-by-step explanation:

The volume of a cone is given by the formula;

V = 1/3 × base area × height

While that of cylinder is given by the formula;

Volume = Base area × height

Since we need the cylinder to fit exactly inside the cone, then it  means that their bases are the same and heights are the same.

Therefore;

The cone has three times less volume compared to the cylinder.

Thus;

Volume of the cylinder = Volume of cone × 3

                                       = 6 In³ × 3

                                       <u>= 18 cubic inches</u>

5 0
3 years ago
Jess walked from her home to her friend’s house and arrived at 4.15. If it took Jess 25 minutes to walk this distance, then what
Airida [17]
D.3:50

Count back 25 minutes from 4:15 and you’ll get 3:50. The numbers on the clock are 5 minutes apart. It should be easy to solve just by using a standard clock
3 0
2 years ago
Read 2 more answers
Please help me with this question, thank you and explain!!!
Novosadov [1.4K]

Total distance from w to z is -10 to 6 = 16

Add the ratios: 2 + 3 + 3 = 8

Divide distance by total ratio: 16/8 = 2

Multiply each ratio by 2 : 2x 2 = 4, 3 x 2 = 6, 3 x 2 = 6

WX is 4 units apart, add 4 to -10 to get x = -6

Answer: A. -6

5 0
2 years ago
The line integral of (2x+9z) ds where the curve is given by the parametric equations x=t, y=t^2, z=t^3 for t between 0 and 1. Pl
Naya [18.7K]
Let r = (t,t^2,t^3)

Then r' = (1, 2t, 3t^2)

General Line integral is:
\int_a^b f(r) |r'| dt

The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector \sqrt{(x')^2 + (y')^2 + (z')^2}

\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt

Fortunately, this simplifies nicely with a 'u' substitution.

Let u = 1+4t^2 +9t^4

du = 8t + 36t^3  dt

\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u}  du \\  \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u}  du \\  \\  \frac{1}{4} \int_0^1 \sqrt{u}  du

After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
=\frac{1}{4} |_0^1 (\frac{2}{3}) (1+4t^2 +9t^4)^{3/2} \\  \\ =\frac{1}{6} (14^{3/2} - 1)
7 0
3 years ago
Solve for x: 4(x+6) = 2-6 (x+3)
laiz [17]

Answer:

x = - 4

Step-by-step explanation:

6 0
3 years ago
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