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Lady_Fox [76]
3 years ago
13

Amber is trying to solve 3x^2-4x=0 using a graphical method. She therefore starts by drawing the graph of y=3x^2 on a grid she n

ow intends to complete her method by drawing a straight line graph in the same grid. complete Ambrrs method to solve 3x^2-4x=0
Mathematics
2 answers:
Delvig [45]3 years ago
6 0
Draw a graph of y=4x, and the x-coordinates of the points of intersection, if I’m not mistaken, are the answers.
AVprozaik [17]3 years ago
5 0

You can rewrite the equation as

3x^2-4x=0 \iff 3x^2=4x

So, you draw the graph of y=3x^2 (the parabola), and then the graph of y=4x (the line), and see where they intersect.

You might be interested in
Ms. Wilson draws a model of the factorization of a polynomial with integer factors. Her model is partially complete.
sergejj [24]
I'll just factor the above equation.

x² + 18x + 80

x² ⇒ x * x
80
can be:
1 x 80
2 x 40
4 x 20
5 x 16
8 x 10  Correct pair

(x+8)(x+10)
x(x+10) +8(x+10) ⇒ x² + 10x + 8x + 80 = x² + 18x + 80

x+8 = 0
x = -8

x+10 = 0
x = -10

x = -8

(-8)² + 18(-8) + 80 = 0
64 - 144 + 80 = 0
144 - 144 = 0
0 = 0

(-10)² + 18(-10) + 80 = 0
100 - 180 + 80 = 0
180 - 180 = 0
0 = 0

I think the algebra tiles will not be a good tool to use to factor the quadratic equation because the equation is not a perfect square quadratic equation.
5 0
3 years ago
Can someone help with this challenging graph ​
-BARSIC- [3]

Answer:

(a) (-∞, -8) (-6, -4) (-2, ∞)

(b) -6, -2

(c) negative

(d) 5

Step-by-step explanation:

(a)  A function is "decreasing" when the y-value decreases as the x-value increases.

⇒ the function is decreasing over these intervals : (-∞, -8) (-6, -4) (-2, ∞)

(b) Local maxima are the points on the function where it reaches a maximum.

⇒ maxima at x = -6 and x = -2

(c) negative

(d) 5

5 0
2 years ago
Find the distance from point A(−1, 7) to the line y=3x. Round your answer to the nearest tenth.
bearhunter [10]

Answer:

<em>The distance from the point to the line is approximately 3.2 units</em>

Step-by-step explanation:

<u>Distance From a Point to a Line</u>

Is the shortest distance from a given point to any point on an infinite straight line. The shortest distance occurs when the segment from the point and the line are perpendiculars.

If the line is given by the equation ax + by + c = 0, where a, b and c are real constants, the distance from the line to a point (x0,y0) is

\displaystyle d= \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}

The line is given by the equation:

y=3x. We need to transform it into the specified form.

Subtracting 3x:

y - 3x = 0

Comparing with the general form of the line, we have

a=-3, b=1, c=0

The point (xo,yo) is (-1,7), thus:

\displaystyle d= \frac{|-3\cdot (-1)+1\cdot 7+0|}{\sqrt{(-3)^2+1^2}}

\displaystyle d= \frac{|3+7|}{\sqrt{9+1}}

\displaystyle d= \frac{|10|}{\sqrt{10}}

\displaystyle d= \frac{10}{\sqrt{10}}

d\approx 3.2

The distance from the point to the line is approximately 3.2 units

4 0
3 years ago
Help please!!!!!!!!!
faltersainse [42]
When the denominator = 0 and the numerator does not, you get a vertical asymptote.  That said a = - 1

If the exponents on the highest variable are the same in the horizontal asymptote is the division of coefficient on the top with the bottom. In plain English what that means is that if you make m = 1 then the coefficients are 2 and 1 (2 in the numerator and 1 in the denominator). If m is any other value greater than 1 there is not a horizontal asymptote.

C <<<====answer.
Third one down.






8 0
3 years ago
Helpppppppppppp plsssssssssssssssssssssssssss!
Goshia [24]

Answer:

<FGH and <JKL are supplementary

<AED and <BEC are complementary

Step-by-step explanation:

For supplementary angles, the sum of the pair of angles must be 180degrees

Given;

<FGH = 109degrees

<JKL = 71degrees

Taking the sum

<FGH + <JKL

= 109 + 71

= 180

Since the sum of the angles is 180, hence <FGH and <JKL are supplementary

Similarly;

Given

<AED = 45degrees

<BEC = 45degrees

Taking the sum

<AED + <BEC

= 45 + 45

=90degrees

Since the sum of complementary angles is 90degrees, hence <AED and <BEC are complementary

8 0
3 years ago
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