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Sunny_sXe [5.5K]
3 years ago
15

What answer to this question?

Mathematics
1 answer:
horrorfan [7]3 years ago
4 0

Answer:

221

Step-by-step explanation:

13 x 17 = 221

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Evaluate the triple integral_E x^6e^y dV where E is bounded by the parabolic cylinder z = 25 - y^2 and the planes z = 0, x = 5,
Nimfa-mama [501]

Nothing crazy here, just a matter of figuring out the limits of integration.

\displaystyle\iiint_Ex^6e^y\,\mathrm dV=\int_{-5}^5\int_{-5}^5\int_0^{25-y^2}x^6e^y\,\mathrm dz\,\mathrm dy\,\mathrm dx

=\displaystyle\int_{-5}^{-5}\int_{-5}^5x^6e^y(25-y^2)\,\mathrm dy\,\mathrm dx

=\displaystyle\left(\int_{-5}^5x^6\,\mathrm dx\right)\left(\int_{-5}^5e^y(25-y^2)\,\mathrm dy\right)=\boxed{\frac{625,000(3+2e^{10})}{7e^5}}

4 0
3 years ago
Ten students begin college at the same time. The probability of graduating in four years is 63%. Which expanded expression shows
Lynna [10]

Answer: the answer is a!


Step-by-step explanation:


5 0
3 years ago
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Ryleigh received a $50 gift card to the frozen yogurt shop for her birthday. The shop sells yogurt sundaes for $4 and yogurt con
timama [110]
The answer is D because the shaded area must go to the origin and it must have a non-dashed line because having 5 sundaes and 10 cones is possible.

9 0
3 years ago
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Find the value of x in the figure below. Assume that the lines are parallel
Andreas93 [3]
C

work:

75 is equal to 2x+15

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5 0
2 years ago
Of the 50 students in an undergraduate statistics class, 60% send email and/or text messages during any given lecture. They have
sasho [114]

Answer:

0.3 or 30%

Step-by-step explanation:

Since no innocent student will ever be caught, the probability that a student sends an email and/or text message during a lecture AND gets caught is given by the product of the probability of a student sending a message (60%) by the probability of the professor catching them (50%) :

P = 0.5*0.6 = 0.3

The probability is 0.3 or 30%.

7 0
3 years ago
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