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3 years ago

Given that (-8,-5) is on the graph of f(x), find

Mathematics

1 answer:

Lemur [1.5K]3 years ago

7 0

Answer:

(-6,-5)

Step-by-step explanation:

Let a function be f(x)

So, f(x-a) would be the same function BUT translated a units to the RIGHT.

So, that changes all x coordinates a units to the right (+a) and all the y-ccordinates remains same.

Our function goes through the transformation f(x-2), which means 2 units to the right, so we "ADD" 2 to all x-coordinates and y remains same.

Thus, the point (-8,-5) would become (-8+2,-5) = (-6,-5)

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Which function describes the arithmetic sequence shown?

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3 years ago

The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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3 years ago

Solve the equation, keeping the value for x as an improper fraction. 23x = − 12x + 5 1.Isolate the variable by adding 12x to bot

kogti [31]

Answer:

$x=\frac{30}{7}$

Step-by-step explanation:

We have been given an equation $\frac{2}{3}x=-\frac{1}{2}x+5$ . We are asked to solve our given equation.

First of all, we will add $\frac{1}{2}x$ on both sides of equation to separate x variable on one side of equation.

$\frac{2}{3}x+\frac{1}{2}x=-\frac{1}{2}x+\frac{1}{2}x+5$

$\frac{2}{3}x+\frac{1}{2}x=5$

Now, we will make a common denominator.

$\frac{2*2}{3*2}x+\frac{1*3}{2*3}x=5$

$\frac{4}{6}x+\frac{3}{6}x=5$

Add numerators:

$\frac{4+3}{6}x=5$

$\frac{7}{6}x=5$

Upon multiply both sides of our equation by $\frac{6}{7}$ , we will get:

$\frac{6}{7}*\frac{7}{6}x=5*\frac{6}{7}$

$x=\frac{5*6}{7}$

$x=\frac{30}{7}$

Therefore, the solution for our given equation is $x=\frac{30}{7}$ .

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3 years ago

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Write and solve an equation based off the verbal phrase: 6 more than x is equal to 33

kenny6666 [7]

Answer:

x + 6 = 33

Step-by-step explanation:

6 more than x means there's an increment in the value of x by addition of 6

x + 6 = 33

This leads to an algebraic linear equation. x is an unknown variable and can be solved for

we subtract 6 from both sides of the equation

x + 6 - 6 = 33 - 6

x = 27

Therefore, we can say 6 more than 27 is 33

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3 years ago