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alexandr402 [8]
3 years ago
6

Complete the table for the given rule y=1/3x i dont have a calculator

Mathematics
1 answer:
Licemer1 [7]3 years ago
5 0

Do up one over 3 and start at 0, that is what I recommend

You might be interested in
Using the quadratic formula to solve 4x2 – 3x + 9 = 2x + 1, what are the values of x?
yaroslaw [1]

Answer:

The value of x is:

x=\dfrac{5\pm \sqrt{103}i}{8}

Step-by-step explanation:

we have to use the quadratic formula to solve for x.

The equation is given as:

4x^2-3x+9=2x+1

which could also be written as:

4x^2-3x+9-2x-1=0\\\\4x^2-3x-2x+9-1=0\\\\\\4x^2-5x+8=0

The quadratic formula for the quadratic equation of the type:

ax^2+bx+c=0 is given as:

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

Here we have:

a=4, b=-5 and c=8.

Hence, by the quadratic formula we have:

x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4\times 8\times 4}}{2\times 4}\\\\x=\dfrac{5\pm \sqrt{25-128}}{8}\\\\\\x=\dfrac{5\pm \sqrt{103}i}{8}

Hence, the value of x is:

x=\dfrac{5\pm \sqrt{103}i}{8}

8 0
3 years ago
Read 2 more answers
7. What is the slope of the line that passes through the pair of points (3, 8) and (9, 5) ? (1 point)
agasfer [191]
7. D
8. D
9. D
10.C

for 7 & 8 you use the equation
\frac{y2 - y1}{x2 - x1}
your points are (3,8) & (9,5)

so you plug the numbers in
y2= 5
y1=8

x2=9
x1=3
you subtract them get a fraction and that is your slope


9 & 10
use the equation
y - y1 = m(x - x1)
plug you numbers in for y1 and x2 m is your slope plug it in and that is your equation


5 0
3 years ago
Read 2 more answers
Explain how the sign of a in the equation y=a(x-h)^2 + k tells you whether the parabola has a minimum or maximum value.
Svet_ta [14]
In short, for a vertical parabola, namely one whose independent variable is on the x-axis, usually is x², if the leading term coefficient is negative, the parabola opens downward, and its peak or vertex is at a maximum, check the picture below at the left-hand-side.

and when the leading term coefficient is positive, the parabola opens upwards, with a minimum, check the picture below at the right-hand-side.

8 0
3 years ago
How does the graph of y=ax^2+c compare to a graph off y=ax^2?
DENIUS [597]
Shifting a graph up by c units involves adding c to the whole function

therefor
y=ax^2+c is c units higher than y=ax^2 (aka, it is shifted c units up)
3 0
3 years ago
Answer to 24 pls and thanks :D ik its super simple but im not very good at this and i forgot how to do it, and you guys get like
iogann1982 [59]
There is no picture given for us to answer the question!
8 0
4 years ago
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