A bell curve ____ A. Normal Distribution B. Negatively Skewed C. Positively Skewed

Suppose f is a function. If 10 = f(-4), find the coordinates of a point on the graph of f?

Rashid [163]

Given:

For a function f, $10 = f(-4)$ .

To find:

The coordinates of a point on the graph of f.

Solution:

We have,

$10 = f(-4)$

It means the value of function f is 10 at -4.

On a coordinate plane x-axis represents the input values and y-axis represents the output values of a function. So,

x-coordinate = input value = -4

y-coordinate = output value = 10

Therefore, the required coordinates of a point on the graph of f are (-4,10).

5 0

4 years ago

Solve using elimination. 8x – 6y = -6 2x – 4y = -14

vovangra [49]

I’m not sure if you want the answer or how to do i’ll just give you both.
multiply the bottom by -4. then it should look like:
8x-6y=-6
-8x+16y=56
then cancel out the x’s and add/subtract the others, giving you: -2y=50. then divide 50 by -2 giving you: y=-25. then find x. plug in y to one of the equations. i usually do the one that hasn’t been messed with. 8x-6(25)=-6. then solve it like a normal two-step equation.
so the answer is: (18,-25)

7 0

3 years ago

What place value is this in 8.7438 the under lined number is 7

MA_775_DIABLO [31]

Answer:

tenths place

Step-by-step explanation:

8 0

3 years ago

PLEASE HELP (30 POINTS) Solve the rational equation x/3 = x^2/x + 5 , and check for extraneous solutions.

anygoal [31]

Option C: $x=0$ and $x=\frac{5}{2}$ are the solutions.

Explanation:

The equation is $\frac{x}{3} =\frac{x^{2} }{x+5}$

We shall determine the value of x, by simplifying the equation.

$\begin{aligned} x(x+5) &=3 x^{2} \\ x^{2}+5 x &=3 x^{2} \\ 2 x^{2}-5 x &=0 \\ x(2 x-5) &=0 \end{aligned}$

Thus, $x=0$ and $x=\frac{5}{2}$ are the solutions.

Now, let us check whether the solutions are extraneous solutions.

Let us substitute $x=0$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}&\frac{0}{3}=\frac{0^{2}}{0+5}\\&0=\frac{0}{5}\\&0=0\end{aligned}$

Thus, both sides of the equation are equal.

Hence $x=0$ is a true solution.

Now, Let us substitute $x=\frac{5}{2}$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}\frac{\left(\frac{5}{2}\right)}{3} &=\frac{\left(\frac{5}{2}\right)^{2}}{\left(\frac{5}{2}\right)+5} \\\frac{5}{6} &=\frac{\left(\frac{25}{4}\right)}{\left(\frac{15}{2}\right)} \\\frac{5}{6} &=\frac{5}{6}\end{aligned}$