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grandymaker [24]
3 years ago
9

The height of a box is 30 inches. What is the height of the box in feet?

Mathematics
2 answers:
IRISSAK [1]3 years ago
6 0

Answer:

2.5 feet

Step-by-step explanation:

KengaRu [80]3 years ago
4 0

Answer:

2.5 feet

Step-by-step explanation:

\frac{30}{12}  = 2.5

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The pot can hold 5.23 quarts

Step-by-step explanation:

  • Step 1: Find the volume of the right cylinder shaped pot.

Diameter of the pot = 8 in

⇒ Radius = 4 in

Height = 6 in

⇒ Volume of the pot = πr²h = 3.14 × 4² × 6

                                   = 301.44 in³

  • Step 2: Convert inches to quarts.

1 in³ = 0.017316 liquid quarts

⇒ 301.44 in³ = 301.44 × 0.017316 = 5.23 quarts

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A cable is 60 decimeters long. How long is the cable in meters?
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Algebra 2B U2 L6 Functions Operations Practice #1-3
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Let R be the region in the first quadrant of the​ xy-plane bounded by the hyperbolas xyequals​1, xyequals9​, and the lines yequa
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Answer:

The area can be written as

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = 0.2274

And the value of it is approximately 1.8117

Step-by-step explanation:

x = u/v

y = uv

Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.

  • x*y = u/v * uv = u², therefore, x*y = 1 when u² = 1. Also x*y = 9 if and only if u² = 9
  • x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4

Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).

Lets compute the partial derivates of x and y over u and v

x_u = 1/v

x_v = u*ln(v)

y_u = v

y_v = u

Therefore, the Jacobian matrix is

\left[\begin{array}{ccc}\frac{1}{v}&u \, ln(v)\\v&u\end{array}\right]

and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))

In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:

\int -v \, ln(v) \, dv = - (\frac{v^2 \, ln(v)}{2} - \int \frac{v^2}{2v} \, dv) = \frac{v^2}{4} - \frac{v^2 \, ln(v)}{2}

Therefore,

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = \int\limits_1^2 (\frac{1}{v} - v \, ln(v) ) (\frac{u^2}{2}\, |_{u=1}^{u=3}) \, dv= \\4* \int\limits_1^2 (\frac{1}{v} - v\,ln(v)) \, dv = 4*(ln(v) + \frac{v^2}{4} - \frac{v^2\,ln(v)}{2} \, |_{v=1}^{v=2}) = 0.2274

4 0
4 years ago
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