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murzikaleks [220]
4 years ago
13

Kaylee found the surface area, in square inches of a rectangular prism by using formula. 2(5×3)+2(5×8)+2(3×8) What is the surfac

e area of the prism, in square inches?

Mathematics
1 answer:
sladkih [1.3K]4 years ago
6 0

Answer:

158

Step-by-step explanation:

Simplify the following:

2×5×3 + 2×5×8 + 2×3×8

5×3 = 15:

2×15 + 2×5×8 + 2×3×8

5×8 = 40:

2×15 + 2×40 + 2×3×8

3×8 = 24:

2×15 + 2×40 + 2×24

2×15 = 30:

30 + 2×40 + 2×24

2×40 = 80:

30 + 80 + 2×24

2×24 = 48:

30 + 80 + 48

| 8 | 0

| 4 | 8

+ | 3 | 0

1 | 5 | 8:

Answer:  158

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vichka [17]

Answer:

  • B. The probability that the student received a grade better than C is 0.45

Step-by-step explanation:

<u>Total number of students:</u>

  • 15 + 30 + 35 + 16 + 4 = 100
<h3>Statements</h3>

A.

  • p(<D) = 4/100 = 0.04
  • 0.04 ≠ 0.2
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B.

  • p(>C) = (30 + 15)/100 = 45/100 = 0.45
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C.

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D.

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6 0
3 years ago
Please help me with this
Kay [80]

Answer:

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Step-by-step explanation:

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7 0
3 years ago
The temperature at 3:00 pm is 27 degrees. The temperature drops 4 degrees every hour for 8 hours. What is the temperature at 11:
Rzqust [24]
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8 0
3 years ago
evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−yj+zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.
meriva

We have

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt

and

\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k

\vec r(t) = \sin(t)\,\vec\imath + \cos(t)\,\vec\jmath + t\,\vec k \implies \dfrac{d\vec r}{dt} = \cos(t) \, \vec\imath - \sin(t) \, \vec\jmath + \vec k

so the line integral is equilvalent to

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt

\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}

\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}

7 0
2 years ago
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