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3 years ago

Which of the following is not an equation of a simple, even polynomial function? y = | x | y = x2 y = x3 y = -x2

Mathematics

2 answers:

BaLLatris [955]3 years ago

6 0

Answer:

The equation $y=x^3$ is not an equation of a simple , even polynomial function.

Step-by-step explanation:

Even function : A function is even when its graph is symmetric with respect to y-axis.

Algebrically , the function f is even if and only if

f(-x)=f(x) for all x in the domain of f.

When the function does not satisfied the above condition then the function is called non even function.

f(x) $\neq$ f(-x)

Now , we check given function is even or not

A. y= $\mid x\mid$

If x is replaced by -x

Then we get the function

f(-x)= $\mid -x \mid$

f(-x)= $\mid x \mid$

Hence, f(-x)=f(x)

Therefore , it is even polynomial function.

B. $y=x^2$

If x is replace by -x

Then we get

f(-x)= $(-x)^2$

f(-x)= $x^2$

Hence, f(-x)=f(x)

Therefore, it is even polynomial function.

C. $y=x^3$

If x is replace by -x

Then we get

f(-x)= $(-x)^3$

f(-x)= $-x^3$

Hence, f(-x) $\neq$ f(x)

Therefore, it is not even polynomial function.

D. $y= -x^2$

If x is replace by -x

Then we get

f(-x)= - $(-x)^2$

f(-x)=- $x^2$

Hence, f(-x)=f(x)

Therefore, it is even polynomial function.

Answer: C. $y=x^3$ is not simple , even polynomial function.

den301095 [7]3 years ago

5 0

Answer:

y = | x |
y = x^3

Step-by-step explanation:

The absolute value function prevents the expression from being a polynomial. The degree of 3 in y^3 is an odd number so that polynomial function will not be even.

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GrogVix [38]

(x-1)^4 = (x-1)²(x-1)²=

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8 0

3 years ago

Mario works at a snack bar near the beach for 2 hours each Saturday. one Saturday, he started with 30 bottles of juice and sold

PSYCHO15rus [73]

Answer:

12 bottles were sold in the second hour.

Step-by-step explanation:

This problem can be solved by implementing an equation in one variable.

The equation will utilize the given values including total bottles of juice, the amount of juice sold in the first hour and amount of juice remaining at the end.

The equation will also consist of the value to be found out, the amount of juice sold in the second hour.

The value to be found out is substituted by a variable.

The equation is then solved by transfer of values and/or variable on the opposite side of the equal sign.

When a value/ variable is transferred on the opposite side of the equal sign, the sign of the value/ variable becomes the opposite.

The negative sign becomes positive and multiplication becomes division, and vice-versa.

1. In the first hour, \frac{3}{15} of all the juice was sold.

\frac{3}{15} of 30 is calculated as

\frac{3}{15} x 30 = 6

6 bottles are sold in the first hour.

2. At the beginning of the second hour, the amount of juice remaining is assumed to be a.

3. As per the question, at the end of second hour, the remaining juice is \frac{6}{15}.

\frac{6}{15} of 30 is calculated as

\frac{6}{15} x 30 = 12

12 bottles are remaining at the end of the second hour.

4. Alternatively, at the end of second hour, 30 - 6 - a of the juice is remaining.

We get, 24 - a bottles are remaining at the end of the second hour.

We equate the above expression with 12, we get the following equation.

24 - a = 12

Next, we transfer a from left side to the right side of the equation.

24 = 12 + a

Next, we transfer 12 from right side to the left side of the equation.

24 - 12 = a

=> 12 = a

a = 12

Hence, 12 bottles were sold in the second hour.

5 0

3 years ago

(Please answer before 5:15 P.M tonight)

taurus [48]

The scale factor of dilation from ABC to $A^{1} B^{1} C^{1}$ is 0.333.

Step-by-step explanation:

Step 1:

First, we plot the coordinates of both triangles.

The coordinates of the triangle $A^{1} B^{1} C^{1}$ are as follows;

$A^{1}$ = (2, 3), $B^{1}$ = (2, 1), and $C^{1}$ = (3, 1).

The coordinates of triangle ABC are;

A = (4, 9), B = (4, 3), and C = (7, 3).

Step 2;

Next, we need to find the distances between at least two coordinates for both triangles.

The distances between vertical lines are found by calculating the difference in y values while for horizontal lines, the distances are found by calculating the difference between the x values.

The distance of $A^{1} B^{1} = 3-1=2$ , the distance of $B^{1} C^{1} = 3-2=1$ .