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fomenos
3 years ago
15

Is this statement true or false

Mathematics
1 answer:
alekssr [168]3 years ago
5 0
The correct answer is true because the origin will always be (0,0)
You might be interested in
Calculate the a) future value of the annuity due, and b) total interest earned. (From Example 2)
vredina [299]

Answer:

  • value: $66,184.15
  • interest: $6,184.15

Step-by-step explanation:

The future value can be computed using the formula for an annuity due. It can also be found using any of a variety of calculators, apps, or spreadsheets.

__

<h3>formula</h3>

The formula for the value of an annuity due with payment P, interest rate r, compounded n times per year for t years is ...

  FV = P(1 +r/n)((1 +r/n)^(nt) -1)/(r/n)

  FV = 5000(1 +0.06/4)((1 +0.06/4)^(4·3) -1)/(0.06/4) ≈ 66,184.148

  FV ≈ 66,184.15

<h3>calculator</h3>

The attached calculator screenshot shows the same result. The calculator needs to have the begin/end flag set to "begin" for the annuity due calculation.

__

<h3>a) </h3>

The future value of the annuity due is $66,184.15.

<h3>b)</h3>

The total interest earned is the difference between the total of deposits and the future value:

  $66,184.15 -(12)(5000) = 6,184.15

A total of $6,184.15 in interest was earned by the annuity.

3 0
2 years ago
4x^4+35x^2-9=0 solve by factoring
Elden [556K]
O si it told me that the solution was -1/2,x2 and it equals =1/2 hope it helps
6 0
3 years ago
4x+5=13 <br> what is x?<br> How do you solve for x?
Mrrafil [7]
4x+5=13 

Start by subtracting 5 from both sides: 

4x+5-5=13-5

4x=8

Divide both sides by 4 to isolate x 

x=2 
6 0
4 years ago
Read 2 more answers
Develop a MATLAB script to generate a 5 panel vertical plot to illustrate how a function changes as the parameters change. On ea
Marat540 [252]

The question is incomplete. Complete question along with Matlab code, explanation, and output results are given below.

Complete Question:

Develop a MATLAB script to generate a 5 panel vertical plot to illustrate how a function changes as the parameters change. On each plot, display the simple sine wave, y(t) = sin(2πt), as a red line. Then, add the following functions to each of the 5-panels as black lines:

y(t) = sin(2πt)  sine function

y1(t)=1+sin(2πt) effect of mean

y2(t)=2sin(2πt) effect of amplitude

y3(t)=sin(4πt) effect of frequency

y4(t)=sin(2πt) - π/4) effect of phase shift

y5(t)=cos(2πt) - π/2) relationship between sine and cosine

Step-by-step explanation:

We are required to show different sinusoidal plots to illustrate the effects of changing mean, amplitude, frequency, phase shift, and relationship of sine and cosine wave.

t=[0:0.01:2*pi] % time vector from 0 to 2pi

y=sin(2*pi*t); % the original sine function  

% effect of mean

y1=1+sin(2*pi*t);  

subplot(5,1,1) % 5 rows, 1 column and last for position

plot(t,y,'k',t,y1,'r')  % this function plots y and y1 with respect to time vector t

% 'k' for black color and 'r' for red color

grid on  

xlabel('time (t)') % x-axis is for time

ylabel('y1(t)')  % y-axis is for function value y(t)

title('effect of mean') % title of the plot

ylim([-3 3]) % limit of y-axis

xlim([0 6]) % limit of x-axis

% effect of amplitude

y2=2*sin(2*pi*t);

subplot(5,1,2)  

plot(t,y,'k',t,y2,'r')  

grid on  

xlabel('time (t)')  

ylabel('y2(t)')  

title('effect of amplitude')  

ylim([-3 3])  

xlim([0 6])  

% effect of frequency

y3=sin(4*pi*t);  

subplot(5,1,3)  

plot(t,y,'k',t,y3,'r')  

grid on  

xlabel('time (t)')  

ylabel('y3(t)')  

title('effect of frequency')  

ylim([-3 3])  

xlim([0 6])  

% effect of phase shift

y4=sin((2*pi*t)-pi/4);  

subplot(5,1,4)  

plot(t,y,'k',t,y4,'r')

grid on  

xlabel('time (t)')  

ylabel('y4(t)')  

title('effect of phase shift')  

ylim([-3 3])  

xlim([0 6])  

% relationship between sine & cosine

y5=cos((2*pi*t)-pi/2);  

subplot(5,1,5)  

plot(t,y,'k',t,y5,'r')  

grid on  

xlabel('time (t)')  

ylabel('y5(t)')  

title('relationship between sine & cosine')  

ylim([-3 3])  

xlim([0 6])  

Output Results:

The first plot shows that sine wave gets shifted to upper side with respect to the original sine wave.

The second plot shows that the amplitude of the sine wave is increased with respect to the original sine wave.

The third plot shows that the frequency of the sine wave is increased. The number of cycles are increased with respect to the original sine wave.

The fourth plot shows there is a phase shift between two waves. The modified sine wave lags the original sine wave by π/4  

The fifth plot shows the relationship between sine and cosine wave.

As we know sin(2πt)=cos(2πt - π/2)

Therefore, both waves are superimposed on each other since they are equal.

3 0
3 years ago
What is the solution to x2 – 35 = 2x ?
Lerok [7]
X=0 because Use the commutative property to reorder the terms , Cancel equal terms on both sides of the equation,
5 0
3 years ago
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