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Pie
3 years ago
7

Find the indicated limit, if it exists. limit of f of x as x approaches 0 where f of x equals 5 x minus 8 when x is less than 0

and the absolute value of the quantity negative 4 minus x when x is greater than or equal to 0
Mathematics
2 answers:
mariarad [96]3 years ago
3 0

Find Lim f(x) as x->0

for

f(x)=5x-8 ......x<0

f(x)=|-4-x|......x &ge; 0


Clearly

f(0)=|-4-0|=|-4|=4 exists.

The left-hand limit, as x-> 0- is f(0-)=5(0)-8=-8

The right hand limit, as x->0+ is f(0+)=|-4-0|=|-4|=4


Since the left and right limits do not both equal f(0), the limit does not exist.


lorasvet [3.4K]3 years ago
3 0

Answer:

The limit does not exists.

Step-by-step explanation:

Given:

f(x) = 5x - 8 ;  x < 0

f(x) = l -4 -x l ; x ≥ 0

Now we have to find the limit when x approaches to 0.

Condition:

If the left hand limit is equal to the right hand limit, then only the limit exists.

Now let's find the left hand limit.

lim         5x - 8

x ---> 0

Now let's apply the limit x = 0, we get

5(0) - 8 = -8

The left hand limit is -8

Now let's find the right hand limit.

lim           l -4 - x l

x ----> 0

Now plug in x = 0 in the above absolute function, we get

l -4 - 0 l = l -4 l = 4 [Absolute value of a negative number is positive]

The right hand limit is 4.

Here we can see that the left hand limit does not equal to the right hand limit.

Therefore, the limit does not exists.

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a rectangular fish tank 60 centimeters by 15 centimeters by 34 centimeters is 1/3 full of water. find the volume of water needed
son4ous [18]

Answer:

20,400\ cm^{3}

Step-by-step explanation:

step 1

Find the volume of the tank

V=(60)(15)(34)=30,600\ cm^{3}

step 2

we know that

If the tank is 1/3 full of water

then

the volume of water required to completely fill the tank is equal to 2/3 of the tank capacity

so

(2/3)30,600=20,400\ cm^{3}

5 0
3 years ago
Please solve the problem ​
jek_recluse [69]

Treat the matrices on the right side of each equation like you would a constant.

Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.

Then you can eliminate <em>Y</em> by taking the sum

4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>

==>   <em>X</em> = (4<em>A</em> + <em>B</em>)/11

Similarly, you can eliminate <em>X</em> by using

-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>

==>   <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11

It follows that

X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}

Similarly, you would find

Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}

You can solve the second system in the same fashion. You would end up with

P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}

3 0
3 years ago
Program: assign03a.m Write a MATLAB script file that uses principal (or amount), rate of interest, and time to compute and displ
DochEvi [55]

Answer:

The code is given in the answer along with the algorithm and explanation. The output and the workspace variables are also attached as screen shots.

Step-by-step explanation:

The basic algorithm of the code is as follows:

  1. Clear the workspace, command window and prepare the format as banking.
  2. Set the constant value of principal amount, this is fixed as 4000 in the code and can be changed.
  3. Ask the user to enter the value of lowest interest rate , highest interest rate , number of interest rate  values and number of years.
  4. Create the interest rate vector.
  5. For each entry of the interest rate vector, calculate the interest.
  6. Display the each entry of interest rate and the simple calculated interest.
  7. Calculate the average rate of interest and the average interest and display them.

The code is here as follows

%% Preparing the formats

clc,clear all;%Clearing command line and workspace

format bank%Formatting

format compact%Formatting

%% Setting the constants

PRINCIPAL = 4000;%Principal amount fixed as 4000.

%% Taking inputs from the user

low_val = input('Enter the low value for the rate range: ');%Lowest value of the interest

high_val = input('Enter the high value for the rate range: ');%Highest value of the interest

rate_vals = input('Enter the number of rate values including high and low values: '); %Total number of rate values

year = input('Enter number of years: ');% Number of years

%% Creating the interest vectors

rate = linspace(low_val,high_val,rate_vals);%Creating the vector

%% Initializing the array

simple_interest=zeros(length(rate),1);%initializing the array to hold the simple interest

%% Displaying the interest rate vectors

fprintf('\t\t Rate(%%)\t\tInterest($)\n');%Displaying the  interest rate and interest heading

%% Calculating the interest vectors

for i = 1:length(rate)

   simple_interest(i) = (rate(i)*year*PRINCIPAL)/100;%Calculating Simple rate of interest for a given rate over the principal amount

fprintf('\t\t %.2f\t\t\t%.2f\n',rate(i),simple_interest(i));% Displaying each entry  interest rate and interest

end

%% Calculating and printing the average interest

fprintf('Average: Rate(%%)\t\tInterest($)\n');%Displaying the heading for average rate and interest

fprintf('\t\t %.2f\t\t\t%.2f\n',mean(rate),mean(simple_interest));%Calculating and displaying the average interest.

4 0
3 years ago
If x &gt; 2, find lx-2l<br><br> Help pleaseee
Lyrx [107]
<h3>Answer:  x-2</h3>

Explanation:

If x > 2, then x-2 > 0 after subtracting 2 from both sides.

Since x-2 is always positive when x > 2, this means the absolute value bars around the x-2 aren't needed. The results of |x-2| and x-2 are perfectly identical.

For example, if we tried something like x = 5, then

  • x-2 = 5-2 = 3
  • |x-2| = |5-2| = |3| = 3

Both outcomes are 3. I'll let you try other x inputs.

So because |x-2| and x-2 are identical, this means |x-2| = x-2 for all x > 2.

In short, we just erase the absolute value bars.

8 0
3 years ago
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Mumz [18]

Answer:

Let's see what to do buddy...

Step-by-step explanation:

2 \frac{3}{5} miles =  \frac{10}{5} +  \frac{3}{5} \: miles =  \frac{13}{5} \: miles \\

So Michael lives 13/5 miles away from the school.

_________________________________

Sarah lives 1/2 as long as Michael.

So the distance from the school of where Sarah's house located is 1/2 distance from the school of where Michael's house located.

Which means :

\frac{13}{5} \div 2 =  \frac{13}{5} \times  \frac{1}{2} =  \frac{13}{10} \\

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

7 0
3 years ago
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