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yaroslaw [1]
3 years ago
12

Which undefined geometric term can be described as a one-dimensional set of points that has no beginning or end

Mathematics
2 answers:
Cerrena [4.2K]3 years ago
6 0

Answer:

A line

Step-by-step explanation:

Komok [63]3 years ago
5 0

Answer:

A line

Step-by-step explanation:

A line is one dimensional

A line has a set of points

A line has no beginning or end, only length

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When –9x” is divided by –3x3, x = 0, the quotient is
iragen [17]

Answer:

0

Step-by-step explanation:

-9 x 0 is 0 and dividing anything with 0 is 0

5 0
3 years ago
Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su
Pachacha [2.7K]
<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

f(x+y)=f(x)+f(y)------(1)  ∀  x,y ∈ R

Now, let us assume f(1)=k

Also,

  • f(0)=0

(  Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0  )

Also,

  • f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1)         ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

  • Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}

Also,

  • when x∈ Q

i.e.  x=\dfrac{p}{q}

Then,

f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq belonging to Q such that:

\to x )

Now, we know that: Q'=R

This means that:

Э α ∈ R

such that Э sequence a_n such that:

a_n\ belongs\ to\ Q

and

a_n\to \alpha

f(a_n)=ka_n

( since a_n belongs to Q )

Let f is continuous at x=α

This means that:

f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha

This means that:

f(\alpha)=k\alpha

                       This means that:

                    f(x)=kx for every x∈ R

4 0
2 years ago
A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int
Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}

By expanding the previous equation:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0

Since there is no accumulation within the tank, expression is simplified to this:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}, where c(0) = 0 \frac{pounds}{gallon}.

\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}

The solution of this equation is:

c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})

The salt concentration after 8 minutes is:

c(8) = 0.166 \frac{pounds}{gallon}

The instantaneous amount of salt in the tank is:

m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds

3 0
3 years ago
One person wants to get a 95% z-confidence interval with a margin of error of at most 15 based on a population standard deviatio
Bess [88]

Answer:

n=(\frac{1.96(60)}{15})^2 =61.46 \approx 62  

So the answer for this case would be n=62 rounded up to the nearest integer  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X represent the sample mean for the sample  

\mu population mean

\sigma=60 represent the population standard deviation  

n represent the sample size (variable of interest)  

Confidence =95% or 0.95

The margin of error is given by this formula:  

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)  

And on this case we have that ME =15, and we are interested in order to find the value of n, if we solve n from equation (1) we got:  

n=(\frac{z_{\alpha/2} \sigma}{ME})^2 (2)  

The critical value for 95% of confidence interval is provided, z_{\alpha/2}=1.96, replacing into formula (2) we got:  

n=(\frac{1.96(60)}{15})^2 =61.46 \approx 62  

So the answer for this case would be n=62 rounded up to the nearest integer  

6 0
3 years ago
What should be done to solve the equation?
stellarik [79]

Answer:

add 17 to both sides

Step-by-step explanation:

as it cancels itself

mark brainliest

5 0
3 years ago
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