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Sloan [31]
3 years ago
15

Which of the binomials below is a factor of this trinomial?

Mathematics
1 answer:
kow [346]3 years ago
8 0

Answer:

x-12

Step-by-step explanation:

Apex

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The distance light travels in one year is known as a light year. A light year is equal to 5.87 × 1012 miles. The closest that Ea
pashok25 [27]

Answer:

6.21\times 10^{-5}\ \text{light years}

Step-by-step explanation:

The distance light travels in one year is known as a light year.

1\ \text{light year}=5.87\times 10^{12}\ \text{miles}

The closest that Earth ever gets to Jupiter is 3.65\times 10^8\ \text{miles}

3.65\times 10^8\ \text{miles}=\dfrac{1}{5.87\times 10^{12}}\times 3.65\times 10^8\ \text{light years}\\\\=6.21\times 10^{-5}\ \text{light years}

Hence, the distance it will take light to travel from Earth to Jupiter at this distance is 6.21\times 10^{-5}\ \text{light years}

4 0
3 years ago
Which figures can the composite figure be broken into? A. a square pyramid and a square prism B. a square pyramid and a cube C.
kifflom [539]

Answer:

c

Step-by-step explanation:

the base dimensons of the bottom portion are not the same therefore it cannot be a cube. so it is c a triangular pyramid and square prism

7 0
3 years ago
Read 2 more answers
Find the volume<br> 12 poir<br> 2.2 cm
Tanzania [10]
Answer: 26.4
To find the volume u multiply 12x2.2=26.4
6 0
3 years ago
PLS HELP ASAP (SHOW WORK) LOTS OF POINTS + give thanks if CORRECT
GuDViN [60]
Multiply by two
2/5 x 2/1 = 4/5
3 0
3 years ago
Let R be the region in the first quadrant of the​ xy-plane bounded by the hyperbolas xyequals​1, xyequals9​, and the lines yequa
Tema [17]

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
3 years ago
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