How to get 1 2/5 from 14/10

1 answer:

5 0

This is a simple simplification problem (ha!).

14/10 is an improper fraction. Since a fraction can also be a division expression, 14 divided by 10 equals 1 with a remainder of 4.

Since we have the answer, we can now produce a mixed fraction of:
4
1 ---
10

Since 4/10 is equal to 2/5,

4 2
1 --- is equal to 1 ---
10 5

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Consider the inverse function. Which conclusions can be drawn about f(x) = x2 + 2? Select three options. f(x) has a limited rang

PolarNik [594]

Answer:

f(x) has a limited range

f(x) has a maximum at the point (0, 2)

f(x) has a y-intercept at the point (0, 2).

Step-by-step explanation:

Given the function;

f(x) = x^2+2

The domain is the value of the input variables for which the function will exist. According to the expression given, the function exists on all real values of x. The same goes with range which deals with the output values. It also exists on all real values from 2 and above.

Hence f(x) have a limited range (since values less than 2 are not included compare to domain that exists on all real values) and does not have a restricted domain.

For the x intercept, x intercept occur at y = 0

substitute y = 0 into the function and get y

if y = f(x)

y = x^2+2

0 = x^2 + 2

x^2 = -2

x = 2i

Hence f(x) does not have an x-intercept of (2, 0)

For the y intercept, y intercept occur at x = 0

substitute x = 0 into the function and get y

if y = f(x)

y = x^2+2

y = 0^2 + 2

y = 2

Hence f(x) has a y-intercept at point (0, 2)

f(x) is at maximum if d(fx))/dx = 0

d(fx))/dx = 2x

since d(fx))/dx = 0

0 = 2x

x = 0

substitute x = 0 into the function

f(x) = x^2 + 2

y = 0^2+2

y = 2

Hence f(x) has a maximum at the point (0, 2)

5 0

3 years ago

Read 2 more answers

Help please I need 5 points total. i need 2 to left of vertex. i need vertex. i need 2 to right of vertex. Please, quickly, I am

zlopas [31]

We are given with a quadratic equation which represents a Parabola , we need to find the vertex of the parabola , But let's recall that , For any quadratic equation of the form ax² + bx + c = 0 , the vertex of the parabola is given by ;