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vlada-n [284]
2 years ago
5

4>x+4 I need to know the x value

Mathematics
1 answer:
AlladinOne [14]2 years ago
6 0

Answer:

x ≤ -1

Step-by-step explanation:

hope this helps

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Lena [83]
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z = 22 - (-9) Solve
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3 years ago
What do you use to repair a TUBA
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Create your own real-world example of a relation that is a function. Domain: The set of Range: The set of
ivanzaharov [21]

Answer:

C(t)=5000 -10t

Step-by-step explanation:

There are many examples in the real world of relationships that are functions.

For example, imagine a tank full of water with a capacity of 5000 liters, this tank has a small hole, by which 10 liters of water are lost every hour.

If we call C the amount of water in the tank as a function of time, then we can write the following equation for C:

C(t)=C_{0}-at\\

Where:

C (t): Amount of water in the tank as a function of time

C_{0}: Initial amount of water in the tank at time t = 0

a: amount of water lost per hour

t: time in hours

Then the equation is:

C(t)=5000 -10t

The graph of C (t) is a line of negative slope. This relation is a function since for each value of t there is a single value of C.

Its domain is the set of all positive real numbers t between [0,500]

Because the time count starts at t = 0 when the tank is full and ends at t = 500 when empty

Its Range is the set of all positive real numbers C between [0,5000] Because the amount of water in the tank can never be less than zero or greater than 5000Litres

7 0
3 years ago
Read 2 more answers
Use the Midpoint Rule with n = 5 to estimate the volume V obtained by rotating about the y-axis the region under the curve y = 1
velikii [3]

Using the shell method, the volume is given exactly by the definite integral,

2\pi\displaystyle\int_0^1x(1+9x^3)\,\mathrm dx

Splitting up the interval [0, 1] into 5 subintervals gives the partition,

[0, 1/5], [1/5, 2/5], [2/5, 3/5], [3/5, 4/5], [4/5, 5]

with left and right endpoints, respectively, for the i-th subinterval

\ell_i=\dfrac{i-1}5

r_i=\dfrac i5

where 1\le i\le5. The midpoint of each subinterval is

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}{10}

Then the Riemann sum approximating the integral above is

2\pi\displaystyle\sum_{i=1}^5m_i(1+9{m_i}^3)\frac{1-0}5

\dfrac{2\pi}5\displaystyle\sum_{i=1}^5\left(\frac{2i-1}{10}+9\left(\frac{2i-1}{10}\right)^4\right)

\dfrac{2\pi}{5\cdot10^4}\displaystyle\sum_{i=1}^5\left(16i^4-32i^3+24i^2+1992i-999\right)=\frac{112,021\pi}{25,000}\approx\boxed{14.08}

(compare to the actual value of the integral of about 14.45)

3 0
3 years ago
3x - 7y = -40 <br><br> 3x - 2y = -4
eimsori [14]

Answer:

What are you solving for?

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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