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HACTEHA [7]
2 years ago
15

SOMEONE PLEASE HELP ME!!!!!

Mathematics
1 answer:
nalin [4]2 years ago
5 0
37 but I’m not sure....
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Which triangle is a 30° -60° -90° triangle?
Fittoniya [83]

Answer:

Acute isosceles triangle

Step-by-step explanation:

because an acute angle is more than 45 but less then 90.

that's all i know sorry!

5 0
3 years ago
How many millimeters are in 9.8 hL
luda_lava [24]

Answer:

980,000 hl

Step-by-step explanation:

There are 980,000 milliliters in 9.8 hectoliters.

5 0
3 years ago
What are the solutions to 4(x+5)^2=4?
aleksandr82 [10.1K]
(X+5)²=4/4
X+5=1,          X+5=-1
X=1-5            X=-1-5
X=-4             X=-6
                      
4 0
3 years ago
Read 2 more answers
Tristan is 22 years younger than Sydney. 8 years ago, Sydney's age was 2 times Tristan's age. How old is Tristan now?
Serjik [45]

Answer:

Sydney is 52 years old, and Tristan is 30.

Step-by-step explanation:

Represent Tristan's age by t and Sydney's by s.

Then s - 8 = 2(t - 8), and t = s - 22

Substituting s - 22 in the first equation, we get:

s - 8 = 2(s - 22 - 8), or

s - 8 = 2s - 60

Solving for s:  -2s + s = -60 + 8, or -s = -52.

Sydney is 52 years old, and Tristan is 30.

4 0
2 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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