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nadezda [96]
3 years ago
9

Please help! I will give you brainliest if it is correct

Mathematics
2 answers:
nlexa [21]3 years ago
6 0

Answer:

B is the correct one. Second.

vodka [1.7K]3 years ago
4 0

Answer:

B

Step-by-step explanation:

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In a dance competition, a participant has to score a total of at least 30 points in the first four rounds combined to move on to
Anvisha [2.4K]

Answer:

5 + 3p ≥ 30

Step-by-step explanation:

"at least" means greater than are equal to. ≥

So we already know he has 5 points, but the other 3 rounds we don't. 5 + 3p

30 is the overall points he has.

5 + 3p ≥ 30

4 0
3 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
What is 23 in simplest form
rodikova [14]
23 or 23/1 or  23.0

Hope this helps ;)
Please don't forget to mark as brainliest answer

5 0
3 years ago
Jesse ears $420 a week, and he works 5 days a week. What is his daily
fredd [130]

Answer:

84 dollars, A

Step-by-step explanation:

420/5 = 84

4 0
3 years ago
Read 2 more answers
459, 450, 441, ...<br> Find the 38th term.
Mademuasel [1]

Answer:

I am pretty sure it is 351

5 0
3 years ago
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