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dusya [7]
3 years ago
10

Successfulness of the Competition policy in South Africa- support and graphs​

Mathematics
1 answer:
jenyasd209 [6]3 years ago
6 0

Answer:

Answer to the following question is a follows;

Step-by-step explanation:

The following are a few examples of how South Africa's competitiveness policy has been successful:

⇒ Consumers or buyers were given a variety of product options as well as competitive prices.

⇒ In 1984, practises like horizontal cooperation and resale price maintenance and control were ruled illegal.

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Pls help me on this question
Step2247 [10]

Answer:H

Step-by-step explanations

Hope this helpes

4 0
3 years ago
A farmer knows that a crop she is growing requires 24 kg of fertilizer throughout the season. She has a measuring flask which wh
Papessa [141]

Answer:

48 times

Step-by-step explanation:

From the above question, we know that:

2 times = 1 kg of fertilizer

A crop requires 24 kg of fertilizer

Hence:

1 kg of fertilizer = 2 times

24 kg of fertilizer = x

Cross Multiply

x = 24 kg × 2 times/1 kg

x = 48 times

She need to fill the flask 48 times to make the required amount of fertilizer for the crop

8 0
3 years ago
What is the solution of 2(4x-2) = 4
JulsSmile [24]

Answer:

Step-by-step explanation:

Divide 2 from both sides.

4x-2 = 4

Add 2 to both sides.

4x = 6

x = 1.5

8 0
3 years ago
Read 2 more answers
x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

3 0
3 years ago
Help!!!!!!!!!!!!!!!!!!!!!!!!
Novay_Z [31]
The answer is x=22 because you just have to add the two equations together and then solve for 8x+4=180
5 0
3 years ago
Read 2 more answers
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