Which statement best describes how to determine wether f(x)=x4-x3 is even a function

Which polygon is a base of the triangular prism?

V125BC [204]

A pyramid or triangular prism is a polyhedron that has a base, which can be any polygon, and three or more triangular faces that meet at a point called the vertex.

Explanation:

These triangular sides are sometimes called the lateral faces to distinguish them from the base.
It is a polyhedron that has a base, which can be any polygon, and three or more triangular faces that meet at a point called the vertex.
The base of the triangular prism is in the shape of a quadrilateral and the lateral face is in the shape of a triangle.

6 0

3 years ago

Read 2 more answers

Antonio finds a pair of skis that cost $350 before tax. Sales tax is 6%. What is the total cost Antonio will have to pay for the

Ksju [112]

Answer:

$371

Step-by-step explanation:

5 0

3 years ago

The function f(x) = 2x + 510 represents the number of calories burned when exercising, where x is the number of hours spent exer

riadik2000 [5.3K]

Answer: 789 calories burned while combining diet with 2 hours of exercise

Step-by-step explanation:

we have that

$f(x)=2x+510$

$g(x)=200x-125$

we know that

$(f+g)(x)=f(x)+g(x)$

substitute

$(f+g)(x)=2x+510+200x-125$

$(f+g)(x)=202x+385$

Find $(f+g)(2)$

For x=2 hours

substitute

$(f+g)(2)=202(2)+385$

$(f+g)(2)=789\ calories$

therefore

The answer is

789 calories burned while combining diet with 2 hours of exercise

8 0

3 years ago

Find a polynomial function of degree 7 with -1 as a zero of multiplicity 3, 0 as a zero of multiplicity 3, and 1 as a zero

trapecia [35]

The polynomial function is P(x) = x^3(x + 1)^3(x -1)

<h3>How to determine the polynomial function?</h3>

The zeros of the polynomial and the multiplicities are given as:

-1 as a zero of multiplicity 3,
0 as a zero of multiplicity 3,
1 as a zero of multiplicity 1

The degree is given as

Degree = 7

The polynomial is represented as:

P(x) = (x - zero)^multiplicity

Using the above format, we have

P(x) = (x + 1)^3(x - 0)^3(x -1)

This gives

P(x) = x^3(x + 1)^3(x -1)

Hence, the polynomial function is P(x) = x^3(x + 1)^3(x -1)

Read more about polynomial function at

brainly.com/question/2833285

#SPJ1

7 0

1 year ago

Can i get some help with #5 please

Sidana [21]

a = 6, is your answer.

Square both sides
9=a−2(6−a)(2a−3)+39=a-2\sqrt{(6-a)(2a-3)}+39=a−2√(6−a)(2a−3)+3

2 .Separate terms with roots from terms without roots
9−a−3=−2(6−a)(2a−3)9-a-3=-2\sqrt{(6-a)(2a-3)}9−a−3=−2√(6−a)(2a−3)

3. Simplify 9−a−39-a-39−a−3 to 6−a6-a6−a
6−a=−2(6−a)(2a−3)6-a=-2\sqrt{(6-a)(2a-3)}6−a=−2√(6−a)(2a−3)

4 .Square both sides
(6−a)2=4(6−a)(2a−3){(6-a)}^{2}=4(6-a)(2a-3)(6−a)2=4(6−a)(2a−3)

5 .Expand
36−12a+a2=48a−72−8a2+12a36-12a+{a}^{2}=48a-72-8{a}^{2}+12a36−12a+a2=48a−72−8a2+12a

6. Simplify 48a−72−8a2+12a48a-72-8{a}^{2}+12a48a−72−8a2+12a to 60a−72−8a260a-72-8{a}^{2}60a−72−8a2
36−12a+a2=60a−72−8a236-12a+{a}^{2}=60a-72-8{a}^{2}36−12a+a2=60a−72−8a2

7. Move all terms to one side
36−12a+a2−60a+72+8a2=036-12a+{a}^{2}-60a+72+8{a}^{2}=036−12a+a2−60a+72+8a2=0

8. Simplify 36−12a+a2−60a+72+8a236-12a+{a}^{2}-60a+72+8{a}^{2}36−12a+a2−60a+72+8a2 to 36−72a+9a2+7236-72a+9{a}^{2}+7236−72a+9a2+72
36−72a+9a2+72=036-72a+9{a}^{2}+72=036−72a+9a2+72=0

9 .Simplify 36−72a+9a2+7236-72a+9{a}^{2}+7236−72a+9a2+72 to −72a+9a2+108-72a+9{a}^{2}+108−72a+9a2+108
−72a+9a2+108=0-72a+9{a}^{2}+108=0−72a+9a2+108=0

10.Factor out the common term 999
−9(8a−a2−12)=0-9(8a-{a}^{2}-12)=0−9(8a−a2−12)=0

11. Factor out the negative sign
−9×−(a2−8a+12)=0-9\times -({a}^{2}-8a+12)=0−9×−(a2−8a+12)=0

12. Divide both sides by −9-9−9
−a2+8a−12=0-{a}^{2}+8a-12=0−a2+8a−12=0

13. Multiply both sides by −1-1−1
a2−8a+12=0{a}^{2}-8a+12=0a2−8a+12=0

14. Factor a2−8a+12{a}^{2}-8a+12a2−8a+12
(a−6)(a−2)=0(a-6)(a-2)=0(a−6)(a−2)=0

15. Solve for aaa
a=6,2a=6,2a=6,2

16 Check solution
When a=2a=2a=2, the original equation −3=6−a−2a−3-3=\sqrt{6-a}-\sqrt{2a-3}−3=√6−a−√2a−3 does not hold true.
We will drop a=2a=2a=2 from the solution set.

17. Therefore,
a=6

5 0

3 years ago