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ELEN [110]
4 years ago
7

What's another name for the x axis and the y axis?

Mathematics
1 answer:
8090 [49]4 years ago
7 0

Answer:

the x axis is also called the axis of abscissas and the Y axis is called the axis of ordinates


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An object moves along a straight line so that at any time t , for 0≤t≤8 , its position is given by x(t)=5+4t−t2 . For what value
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It’s ccccccccccccccccccccccc
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Which equation shows y=−12x+4 in standard form?
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The answer is the first option

6 0
3 years ago
Enter the correct answer in the box. solve the equation x2 − 16x 54 = 0 by completing the square. fill in the values of a and b
poizon [28]

The roots of the given polynomials exist  $x=8+\sqrt{10}$, and $x=8-\sqrt{10}$.

<h3>What is the formula of the quadratic equation?</h3>

For a quadratic equation of the form $a x^{2}+b x+c=0$ the solutions are

$x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Therefore by using the formula we have

$x^{2}-16 x+54=0$$

Let, a = 1, b = -16 and c = 54

Substitute the values in the above equation, and we get

$x_{1,2}=\frac{-(-16) \pm \sqrt{(-16)^{2}-4 \cdot 1 \cdot 54}}{2 \cdot 1}$$

simplifying the equation, we get

$&x_{1,2}=\frac{-(-16) \pm 2 \sqrt{10}}{2 \cdot 1} \\

$&x_{1}=\frac{-(-16)+2 \sqrt{10}}{2 \cdot 1}, x_{2}=\frac{-(-16)-2 \sqrt{10}}{2 \cdot 1} \\

$&x=8+\sqrt{10}, x=8-\sqrt{10}

Therefore, the roots of the given polynomials are $x=8+\sqrt{10}$, and

$x=8-\sqrt{10}$.

To learn more about quadratic equations refer to:

brainly.com/question/1214333

#SPJ4

3 0
2 years ago
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Can someone help with numbers 1 and 2 for both of them tell me the answer and how you get it please I need the two of them do no
algol [13]

Answer:

look it up

Step-by-step explanation:

L.O.O.K   I.T      U.P.

5 0
3 years ago
Please help on this one? :)
kirill [66]

Answer:B,C,D


Step-by-step explanation:B- The point is over 1 off the y-axis

C-They are the points the line crosses the x-axis

D- There is no lower point


7 0
4 years ago
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