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baherus [9]
3 years ago
14

(1) Ax + By = C (2) Dx + Ey = F Jim is trying to solve the system of equations. He begins by multiplying equation (1) by D and e

quation (2) by A. Before he can continue, his friend Angela comes by and says, "No, you should have multiplied equation (1) by E and equation (2) by B. You're going to get the wrong answer." Is Angela right? Why or why not?
Mathematics
2 answers:
attashe74 [19]3 years ago
8 0

Answer:

No, Angela is not correct.

As the method Jim was performing will also lead to the solution.

Step-by-step explanation:

We are given first equation as:

                  Ax + By = C

Second equation is:

                   Dx + Ey = F

Jim solved the equation as:

He begins by multiplying equation (1) by D and equation (2) by A.

and so by subtracting both the equations he will obtain a value of y.

and then put the y-value in any of the given two equations to obtain the value of x.

Angela Method:

you should have multiplied equation (1) by E and equation (2) by B.

and when she will subtract both the equations she will get the value of x first and then when she will put the value of x in any of the given equation she will obtain the value of y.

But both will get a value for x and y.

Hence, the method Jim was performing was also correct.

Butoxors [25]3 years ago
6 0

Answer:

C

Step-by-step explanation:  

No; Jim is trying to eliminate x, while she is trying to eliminate y. Both ways will work.

You might be interested in
The average (arithmetic mean) of five different positive integers is 30. What is the greatest possible value of one of these int
geniusboy [140]

Answer:

The Greatest possible value of one of the integer is 140.

Step-by-step explanation:

Given:  The average (arithmetic mean) of five different positive integers is 30.

To find:  What is the greatest possible value of one of these integers?

Explanation: we are given that the arithmetic mean of five  positive integer

                   is 30.

Let the five positive integers are A,B,C,D,E

The arithmetic mean :

                   \frac{A+B+C+D+E}{5} =30.

On multiplying both side by 5.

                       A+B+C+D+E =150.

The least values of A ,B,C and D can be 1 ,2,3,4.

  Then :  1 +2+3+4+E =150

On simplification  10+E =150

On subtraction both side by 10 we get

                      E = 140 .

Therefore, the Greatest possible value of one of the integer is 140.


6 0
3 years ago
Can someone help me please???​
Karolina [17]

Answer:

Altitude of Equilateral Triangle h = (1/2) * √3 * a. Angles of Equilateral Triangle: A = B = C = 60° Sides of Equilateral Triangle: a = b = c.

Hope this helped!

If you would like me to simplify it a little let me know.

8 0
3 years ago
Help please last one
Monica [59]
Y=1/2x+8


Step by step explanation This is how I got the answer to your question and I gave you the solution I hope this helps you out
8 0
2 years ago
Park Rangers at the North Rim of the Grand Canyon recorded the amounts of rainfall over 12 months. What was the total amount of
myrzilka [38]

Answer:

65.532cm of rainfall over 12 months

Step-by-step explanation

5 0
3 years ago
An article presents voltage measurements for a sample of 66 industrial networks in Estonia. Assume the rated voltage for these n
kramer

Answer:

We accept the null hypothesis and conclude that voltage for these networks is 232 V.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 232 V

Sample mean, \bar{x} = 231.5 V

Sample size, n = 66

Sample standard deviation, s = 2.19 V

Alpha, α = 0.05

First, we design the null and the alternate hypothesis

H_{0}: \mu = 232\\H_A: \mu \neq 232

We use Two-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n-1}} } Putting all the values, we have

t_{stat} = \displaystyle\frac{231.5- 232}{\frac{2.19}{\sqrt{66}} } = -1.8548

Now,

t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = \pm 1.9971

Since,              

|t_{stat}| > |t_{critical}|

We accept the null hypothesis and conclude that voltage for these networks is 232 V.

4 0
3 years ago
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