All categories

2 years ago

What is the solution to this equation?

Mathematics

1 answer:

exis [7]2 years ago

7 0

I need help on that as well

You might be interested in

Find each variable please!!!

elena-14-01-66 [18.8K]

Answer:

x = 6.5

Step-by-step explanation:

By property of intersecting secants outside of circle.

$6(6 + x) = 5(5 + 10) \\ 36 + 6x = 5 \times 15 \\ 6x = 75 - 36 \\ 6x = 39 \\ x = \frac{39}{6} \\ x = \frac{13}{2} \\ x = 6.5$

8 0

3 years ago

x and y are integers between 0 and 10 that satisfy the equation shown below. x2=y⎯⎯√+3 What values for x and y satisfy the equat

8_murik_8 [283]

7 0

3 years ago

Identify the Quadrant where the point is located.

ValentinkaMS [17]

Answer:

1 ko second quadrant and 2 ko chai fourth quadrant

Step-by-step explanation:

plz mark me as brainest answer

8 0

3 years ago

Erik and Caleb were trying to solve the equation: 0=(3x+2)(x-4) Erik said, "The right-hand side is factored, so I'll use the zer

Mumz [18]

Answer:

C) Both

Step-by-step explanation:

The given equation is:

$0=(3x+2)(x-4)$

To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.

Using this property:

$(3x+2) = 0 \Rightarrow \bold{x = -\frac{2}{3}}\\(x-4) = 0 \Rightarrow \bold{x = 4}$

So, Erik's solution strategy would work.

Now, let us discuss about Caleb's solution strategy:

Multiply $(3x+2)(x-4)$ i.e. $3x^2-12x+2x-8$ = $3x^2-10x-8$

So, the equation becomes:

$0=3x^2-10x-8$

Comparing this equation to standard quadratic equation:

$ax^2+bx+c=0$

a = 3, b = -10, c = -8

So, this can be solved using the quadratic formula.

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

$x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times3 \times (-8)}}{2\times 3}\\x=\dfrac{-(-10)\pm\sqrt{196}}{6}\\x=\dfrac{10\pm14}{6} \\\Rightarrow x= 4, -\dfrac{2}{3}$