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Brut [27]
2 years ago
10

How many 3/8’s are in 6

Mathematics
1 answer:
Agata [3.3K]2 years ago
8 0

\huge\mathfrak\pink{Answer}

there are 16 3/8s in 6

\huge\mathcal\green{hope \: this \:may }\huge\mathcal\pink{be \: helpful}

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PLS HELP WILL GIVE BRAINLIEST ANSWER
lubasha [3.4K]

Answer:

56°

Step-by-step explanation:

First, the measure of angle GHE is equal to DFH. This means that you can set 98-6x=63-x, which becomes 5x=35 so x=7. Substitute 7 for x and you'll get DFH =63-7, which is 56.

8 0
2 years ago
What is the sum? <br> Thank you!
timurjin [86]

Answer:

50 is the sum

Step-by-step explanation:

-10+20=10

10+-40=-30

-30+80=50

5 0
2 years ago
26 x 5/12 explain in steps
hoa [83]
First simplify: 26 x 5/12 would equal to 13 x 5/6 because we divided it 2, then do the equation, which would now be 65/6, 13 x 5 and 1 x 6 so now we can make it into mixed numbers, so 10 5/6

ANSWER: 65/6 or 10 5/6
8 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
Square root 11 +3 square root 11
Anastasy [175]

Answer:

13.2664991614

3 0
2 years ago
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