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3 years ago

How many real number solutions does the equation have 0=5x^2+2x-12

2 answers:

7 0

Rearrange the equation to standard form of a quadratic equation (ax^2+bx+c=0) by switching sides: 5x^2+2x-12=0. Now, use the quadratic equation formula to solve. You should come out with x_1=sqrt61-1/5 and x_2=-1+sqrt61/5. Thus, your answer is B, or two solutions.

aniked [119]3 years ago

3 0

Answer:

A. two solutions

Step-by-step explanation:

To find the number of real solutions the equation is having, we need to solve first;

0=5x²+2x-12

This equation can be re-written as;

5x²+2x-12 = 0

We will use the formula method to solve

Using formula method;

x = -b ±√b² - 4ac / 2a

From the equation given,

a = 5 b = 2 and c=-12

x = -2 ± √2² - 4(5)(-12) / 2(5)

x = -2 ± √4 + 240 / 10

x = -2/10 ± √244 /10

x = $\frac{-1}{5}$ ± 1.56

x = - 0.2 ± 1.56

Either x = -0.2 + 1.56 or x = -0.2 - 1.56

Either x = 1.36 or x = -1.76

Therfore, this equation have two real number solutions

You might be interested in

Ayden wants to sell a total of $150 worth of candy bars for his school fundraiser. Each candy bar costs $1.50. How many candy ba

worty [1.4K]

Answer:

100 Candy Bars

Step-by-step explanation:

Hey!

==================================================================

Ayden wants to sell a total of $150 worth of candy bars for his school fundraiser. Each candy bar costs $1.50. How many candy bars will he have to sell in order to meet his goal?

--------------------------------------------------------------------------------------------------------------

We can create an algebraic Equation to solve this.

- We know that Each Candy Bar costs $1.50.

-We multiply the Amount of Candy Bars with the price.

-We know the profit should be $150

⇒ 1.50x = 150

-Divide by 1.50 on both sides

⇒ 1.50x/1.50 = 150/1.50

-Simplify

⇒ x = 100

==================================================================

Hope I Helped, Feel free to ask any questions to clarify :)

Have a great day!

More Love, More Peace, Less Hate.

-Aadi x

4 0

3 years ago

Graph ΔABC and its image after a rotation of 180º about the origin.

ValentinkaMS [17]

Answer:

The vertices of image are A'(0,0), B'(-1,-5) and C'(4,-5). The graph of image and preimage is shown below.

Step-by-step explanation:

From the given figure it is noticed that the vertices of triangle ABC are A(0,0), B(1,5) and C(-4,5).

If a figure rotated at 180º about the origin, then

$P(x,y)\rightarrow P'(-x,-y)$

The vertices of image are

$A(0,0)\rightarrow A'(0,0)$

$B(1,5)\rightarrow B'(-1,-5)$

$C(-4,5)\rightarrow C'(4,-5)$

Therefore the vertices of image are A'(0,0), B'(-1,-5) and C'(4,-5).

The graph of image and preimage is shown below.

3 0

3 years ago

What is an equation of the line that passes through the point (-6,-8) and is parallel to the line x-2y=6?

vfiekz [6]

Answer:

$y=\frac{1}{2}x-5$

Step-by-step explanation:

Hi there!

What we need to know:

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept (the value of y when the line crosses the y-axis)
Parallel lines always have the same slope and different y-intercepts

1) Determine the slope (m)

$x-2y=6$

Rearrange this equation into slope-intercept form (this will help us find the slope)

Subtract x from both sides

$x-2y-x=6-x\\-2y=-x+6$

Divide both sides by -2

$y=\frac{1}{2} x-3$

Now, we can identify clearly that the slope of the given line is $\frac{1}{2}$ since it's in the place of m. Because parallel lines always have the same slopes, the line we're currently solving for would therefore have a slope of $\frac{1}{2}$ as well. Plug this into $y=mx+b$ :