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Minchanka [31]
3 years ago
15

Y

Mathematics
2 answers:
baherus [9]3 years ago
8 0
It’s c for sure only because I had it on a test
den301095 [7]3 years ago
3 0

Answer:

a bc I know it and I got it ⏯️ ght yes sir

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If three less than one half a number is equal to one-third of the same number, find the number​
Anastasy [175]

Answer: The number would be 18

Step-by-step explanation:

18/2=9 9-3=6 6*3=18

7 0
3 years ago
Which set of numbers gives the minimum, lower quartile, median, upper quartile, and maximum, in that order, of this box plot?
Illusion [34]
I think it is E .I am thankfull in helping you
4 0
3 years ago
Gareth buys two oranges. He pays with a £1 coin and gets 52p change. Work out the cost of one orange.
zalisa [80]

Answer:

24p

Step-by-step explanation:

£1 - £0.52 = £0.48 ← cost of 2 oranges

cost of 1 orange = £0.48 ÷ 2 = £0.24

3 0
2 years ago
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
498 divided by 13 full Equation
Rudiy27

Answer:

38.3

Step-by-step explanation:

hope I helped

6 0
2 years ago
Read 2 more answers
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