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Serjik [45]
2 years ago
15

Question: Find Angle S SOMEONE PLEASE HELP WILL GIVE BRAINLEST!

Mathematics
1 answer:
fenix001 [56]2 years ago
4 0

Answer: 45

Step-by-step explanation:

S is 45 (labelled on the diagram)

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Given the linear equation 2x + y = 6, perform the necessary operations to put the equation into the proper general form. Explain
likoan [24]

Step-by-step explanation:

The general equation is y = mx + c...

The given equation is 2x + y = 6

Firstly, move everything on the left except the y to the right.i.e.

y = 6 - 2x

Secondly, rearrange the values on the right to look just like the general equation.

y = -2x + 6

The equation is now in proper general form because it matches the format laid down.

3 0
3 years ago
Write the equation of the line in fully simplified slope-intercept form.​
Aneli [31]

Answer:

y = -\frac{1}{5}x -1

Step-by-step explanation:

Given

The attached graph

Required

Determine the line equation

First, list out two points from the graph

(x_1,y_1) = (-5,0)

(x_2,y_2) = (0,-1)

Next, calculate the slope (m)

m = \frac{y_2 - y_1}{x_2 - x_1}

m = \frac{-1-0}{0 -(-5)}

m = \frac{-1}{0 +5}

m = -\frac{1}{5}

The equation in slope intercept form is:

y = m(x - x_1) + y_1

This gives:

y = -\frac{1}{5}(x - (-5)) + 0

y = -\frac{1}{5}(x +5)

Open bracket

y = -\frac{1}{5}x -1

4 0
2 years ago
Does increase in sample size increases degrees of freedom for t-tests.
Elena-2011 [213]
Basically degrees of freedom are related to sample size (n-1). If the df increases, it also stands that the sample size is increasing; the graph of the t-distribution will have skinnier tails, pushing the critical value towards the mean.
3 0
2 years ago
What is the parent function of the graph?
Alik [6]

Answer:

y = |x| – 4

Step-by-step explanation:

If we substitute x as 0, we get -4 therefore this is the answer.

8 0
3 years ago
Find the distance from the origin to the graph of 7x+9y+11=0
Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

Solving for (d(x)^2)'=0, you find a critical point of x=-\dfrac{77}{130}.

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0

so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
4 0
3 years ago
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