Let side be n
#1
#2
#3
3y-6=3x and y+2x=5 solve the system of linear equations by graphing PLEASE SHOW WORK
1 answer:
Answer:
(x, y) = (1, 3)
Step-by-step explanation:
The "work" of solving by graphing is typing the equations into a graphing calculator and highlighting the point of intersection.
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If you're graphing these by hand, there are a couple of strategies you can use. One is to make use of the intercepts of the line. For each of these, you need to rearrange the equation to a suitable form.
<u>3y -6 = 3x</u>
To put this into "intercept form", put the variable terms on one side of the equal sign and the constant on the other. Then divide by the constant and express the resulting variable coefficients as denominators.
3x -3y = -6
x/(-6/3) +y/(-6/-3) = 1
x/(-2) + y/2 = 1 . . . . . . The x-intercept is -2; the y-intercept is 2.
Plot the intercept points and draw a line through them.
<u>y +2x = 5</u>
The intercepts for this one are found by dividing by 5:
x/(5/2) + y/5 = 1 . . . . . . The x-intercept is 2.5; the y-intercept is 5.
The other strategy you can use for this one is to rearrange the equation to slope-intercept form and make use of those numbers.
y = -2x + 5
The y-intercept is 5 (as we already found above), and the slope is -2. That means the line drops 2 units for each unit it goes to the right. Apart from the y-intercept, other points on this line will be (1, 3), (2, 1), (3, -1).
<u>Solution</u>
When you plot these lines, you find they intersect at (1, 3). That is the solution to the system of equations, the values of x and y that satisfy both equations.
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Answer:
it must also have the root : - 6i
Step-by-step explanation:
If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.
This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:
where the imaginary unit has disappeared, making the expression real.
So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)
Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.
Answer:
z = -1.645
Step-by-step explanation:
b) n =25, M = 4.01 , \mu = 4.22 , \sigma = 0.60 , \alpha= 0.05
The hypothesis are given by,
H0 : \mu\geq 4.22 v/s H1 : \mu < 4.22
The test statistic is given by,
Check attachment for the formula that should br here.
z = \frac{4.01- 4.22 }{0.60 /\sqrt{25}}
= -1.75
The critical value of z = -1.645
The calculated value z > The critical value of z
Hence we reject null hypothesis.
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