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katrin [286]
3 years ago
7

What is expanded form 12.38

Mathematics
2 answers:
Aleks [24]3 years ago
6 0
Expanded form = Twelve and thirty- eight hundredths.<span>
</span>
Kruka [31]3 years ago
5 0
Twelve and thirty- eight hundredths.
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(4+5)-2 thxs for answering!!!!!!!!!!!!!!
Arisa [49]

Answer: 7

Step-by-step explanation:

(4+5)-2

You can either remove brackets and just do a little addition and subtraction.

4+5-2

9-2\\7

So the answer is 7.

Or just finish things inside the brackets first.

(4+5)-2\\9-2\\7

So the answer is 7.

3 0
4 years ago
What is the answer of 8-15=
kicyunya [14]

Answer:

-7

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
angle A and angle B are supplementary angles. If m angle A=(7x+14)^ and m angle B=(5x-26)^ , then find the measure of angle B .
Aleksandr [31]

Answer:

35x^2 - 112x - 364

it's probably wrong since I've only expanded the brackets

5 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
4 years ago
NEED HELP ASAP! The graph gx is a translation of the function fx=x2. The vertex of gx is located 5 units above and 7 units to th
OLga [1]

Answer:

g(x) is option 1 or g(x) = (x + 7)^2 + 5

A quadratic equation has the general form of:

y=ax² +bx + c

It can be

converted to the vertex form in order to determine the vertex of the parabola.

It has the standard form of:y =a(x+h)² + k

where hand k represents the vertex, h represent the point in the x axis and k is thepoint in the y axis. Therefore, from the details given in the problem, the equation that represents

g(x) is option 1 or g(x) = (x + 7)^2 + 5

8 0
3 years ago
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