All categories

3 years ago

Which statement is true regarding the graphed functions?

Mathematics

1 answer:

harina [27]3 years ago

6 0

<h3>Answer: Choice A. f(0) = g(0)</h3>

This is because the red and blue lines cross at (0,-2). We don't really need to worry about the y coordinate here. For this problem, all we care about is the x coordinate, which is x = 0.

When x = 0, the outputs of each function f(x) and g(x) are both the same value. So that's why we can say f(0) = g(0). It's the same as saying f(x) = g(x) has the solution x = 0.

You might be interested in

Two piecewise functions are shown below. What is the value of

gizmo_the_mogwai [7]

Step-by-step explanation:

I am not sure what your problem here is.

you understand the inequality signs ?

anyway, to get

6×f(-2) + 3×g(1)

we can calculate every part of the expression separately, and then combine all the results into one final result.

f(-2)

we look at the definition.

into what category is -2 falling ? the one with x<-2, or the one with x>=-2 ?

is -2 < -2 ? no.

is -2 >= -2 ? yes, because -2 = -2. therefore, it is also >= -2.

so, we have to use

1/3 x³

for x = -2 that is

1/3 × (-2)³ = 1/3 × -8 = -8/3

g(1)

again, we look at the definition.

into what category is 1 falling ? the one with x > 2 ? or the one with x <= 1 ?

is 1 > 2 ? no.

is 1 <= 1 ? yes, because 1=1. therefore it is also <= 1.

so we have to use

2×|x - 1| + 3

for x = 1 we get

2×0 + 3 = 3

6×f(-2) = 6 × -8/3 = 2× -8 = -16

3×g(1) = 3× 3 = 9

and so in total we get

6×f(-2) + 3×g(1) = -16 + 9 = -7

3 0

2 years ago

Write a quadratic equation with the given roots. Write the equation in the form of ax^2+bx+c=0 where a b and c are integers

Bas_tet [7]

You haven't provided the required roots, but I can tell you how to do this kind of exercises in general.

If the $x^2$ coefficient is 1, i.e. the equation is written like $x^2+bx+c=0$ , then you can say the following about the coefficients b and c:

$b$ is the opposite of the sum of the roots
$c$ is the multiplication of the roots.

So, for example, if we want an equation whose roots are 4 and -2, we have:

$4+(-2) = 4-2 = 2 \implies b = -2$
$4 \cdot (-2) = -8 \implies c = -8$

So, the equation is $x^2-2x-8=0$

If your roots are rational, you can work like this: suppose you want an equation with roots 3/4 and 1/2. You have:

$\dfrac{3}{4}+\dfrac{1}{2} = \dfrac{3}{4}+\dfrac{2}{4} = \dfrac{5}{4} \implies b = -\dfrac{5}{4}$
$\dfrac{3}{4} \cdot \dfrac{1}{2} = \dfrac{3}{8} \implies c = \dfrac{3}{8}$