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Marizza181 [45]
3 years ago
7

Please help me with number 2 :)

Mathematics
1 answer:
Tcecarenko [31]3 years ago
7 0

Aubra because you always round to the greater value if the number is 5-9

You only round down if the number at the end is 0-4

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70 points
Verizon [17]

Answer:

500

All of the above

paying your balances off in full

make credit card purchases that you know you can pay back within a month

0% interest for 1 year and 12% interest after that

0% interest for 1 year and 12% interest after that

paying your balances off in full

all of the above

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What is the first term of the quotation of the following division problem?
sveticcg [70]

Answer:

I believe it's D.

Step-by-step explanation:

x^2 times x

Hope This Helps!       Have A Nice Day!!

5 0
3 years ago
If a bus traveled 160 miles in 4 hours, what was the average speed in miles per hour
sashaice [31]
160/4 = 40 mph so that would be the answer simply divide
5 0
2 years ago
A cup of coffee which is initially at a temperature of 154∘F is placed in a room which is at a constant temperature of 71∘F. In
Lesechka [4]
Newton's law of cooling is
k (t₁ - t₂) = -ln (T₁ - T∞ / T₂ - T<span>∞)

Use the two data points in the given to find k.

t = 0, T = 154
t = 10, T = 133

Solution:

k(0 - 10) = - ln (154 - 71/ 133 - 71)
-10k = -0.291716
k = 0.02917 or 0.0292

So now find t when T = 100
0.2917 * ( 0 - t) = -ln (154 - 71 / 100 -71)
- 0.02917t = - 1.0515
t = 36. 05 minutes
to the nearest minute t = 36</span>
6 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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