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3 years ago

Factor completely. 4x^2-2x-12

1 answer:

4 0

4x^2-2x-12=2x^2-x-6(On dividing by 2) Or, 2x^2-x-6=0 Or, 2x^2-4x+3x-6=0 Or, 2x(x-2)+3(x-2)=0 Or, the factors are: (2x+3)(x-2)

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Help! Thanks I am not smart.

Gnom [1K]

Open the numbers for easy grasp by just knowing the decimal position but also you can read directly from the numbers and checking their exponents.

1.(Jurassic) 208000000

2.(Silurian) 438000000

3.(Tertiary) 66400000

4.(Triassic) 245000000

245000000Silurian, Triassic, Jurassic and Tertiary.

245000000Silurian, Triassic, Jurassic and Tertiary. And yes you are smart! Don't sell yourself short!

245000000Silurian, Triassic, Jurassic and Tertiary. And yes you are smart! Don't sell yourself short! All the best!

4 0

3 years ago

How do I factor this? X^2-12x+36

LiRa [457]

$x^2-12x+36 =\\
(x-6)^2$

6 0

4 years ago

Read 2 more answers

What is the measure of ACD? - 100 - 130 - 50 - 80

malfutka [58]

Answer:

130°

Step-by-step explanation:

80+50= 130

180-130=50

180-50=130°

3 0

3 years ago

The variables x and y vary directly. Use the values to find the constant of proportionality, k . Then write an equation that rel

Vaselesa [24]

Answer:

k = 9/8

y = 9/8x

Step-by-step explanation:

45/40 = 9/8

To check:

40 * 9/8 = 45

Hope this helps!

7 0

3 years ago

WILL MAKE BRAINLIEST!!!!Complete the system of equations with a linear equation so that the graphs of the equations intersect at

Nina [5.8K]

First, we have to find x- and y-intercepts of given equation.

1. y-intercept is a point (0, y) so substitute x = 0 and calculate:

$y=x^3-3x^2+x-3\\\\y= 0^3-3\cdot0^2+0-3\\\\y=0-0+0-3\\\\\boxed{y=-3}$

y-intercept = (0, -3)

2. x-intercept is a point (or points) (x, 0), so when we substitute y = 0 there will be:

$0= x^3-3x^2+x-3\\\\0=x^2(x-3)+(x-3)\\\\0=(x^2+1)(x-3)\\\\\\x^2+1=0\qquad\vee\qquad x-3=0\\\\x^2+1\text{ always}\ \textgreater \ 0\qquad\vee\qquad x=3\\\\\boxed{x=3}$

x-intercept = (3, 0)

Now, when we have x- and y-intercepts, we can write an equation of a line (graph of the linear function) that passes through this two points:

$A=(3,0)\qquad\qquad B=(0,-3)\\\\\\
y-y_A=\dfrac{y_B-y_A}{x_B-x_A}(x-x_A)\\\\\\y-0=\dfrac{-3-0}{0-3}(x-3)\\\\\\y=\dfrac{-3}{-3}(x-3)\\\\\\y=1(x-3)\\\\\\\boxed{y=x-3}$

Our linear equation is: y = x - 3

5 0

4 years ago