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omeli [17]
2 years ago
14

A. (-3,7) B. (-3,-7) C. (3,-7) D. (3,7)

Mathematics
1 answer:
Rufina [12.5K]2 years ago
8 0
The answer is C. (3,-7)
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100min to 2h as a ratio
Usimov [2.4K]

Answer:

100:2 or 50:1

Step-by-step explanation:

Hope this helps!

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4 0
3 years ago
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Please help this is due in 1 hr.
defon
Answer:
Mr.Brown received $5,602.5 last month

Step-by-step explanation:
Mr.Brown already has $5,600 and has 2.5. 2.5 as a fraction is 2 1/2, but it is easier having it as a decimal, so since it’s already a decimal all you need to do it add 5,600 and 2.5(5,600 + 2.5).
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2 years ago
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ANSWER FOR 30 POINTS!
Darya [45]

Answer:

48

Step-by-step explanation:

27+36-15=63-15=48

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3 years ago
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Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term x^2, which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form f(x)=ax^2+bx+c is given by the expression: x_v=\frac{-b}{2a}

Since in our case a=1 and b=5, we get that the x-position of the vertex is: x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the x_v=-\frac{5}{2}. Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

5 0
3 years ago
Determine the standard variation of the data below. (1, 2, 3, 4, 5)
Ne4ueva [31]
Calculate for the mean/ average of the given numbers:
 
                             μ = (1 + 2 + 3 + 4 + 5) / 5 = 3

Then, we calculate for the summation of the squares of differences of these numbers from the mean, S
 
                             S  = (1 - 3)² + (2 - 3)² + (3 - 3)² + (4 - 3)² + (5 - 3)²
                                S = 10

Divide this summation by the number of items and take the square root of the result to get the standard deviation.

                              SD = sqrt (10 / 5) = sqrt 2  
                                    SD = 1.41

Thus, the standard deviation of the given is equal to 1.41. 
              
                            
8 0
3 years ago
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