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Vikki [24]
4 years ago
11

Is this odd f(x)=X^3-X^2

Mathematics
2 answers:
Oksanka [162]4 years ago
8 0
No it is not I believe
Dmitrij [34]4 years ago
4 0

Is this odd f(x)=X^3-X^2

No.

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Find the area of a circle with a circumference of 6.28 unitesz
Anni [7]

Answer:

3.14 square units

Step-by-step explanation:

Circumference of a circle = 2πr

2πr = 6.28

r = 6.28/2π

r = 3.142/π

Area of a circle = πr^2

slot in the value of r

Area of the circle = π( 3.142/π)^2

               "              = π(9.87/ π^2)

                "             = 9.87/π

π = 3.14

Area of a circle = 9.867/ (3.14)

              "            = 3.14 square units

                     

8 0
3 years ago
A person invests 4500 dollars in a bank. The bank pays 6.25% interest compounded
Len [333]

Answer:

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Solve for x in the diagram below
andre [41]

Answer:

x= 20

Step-by-step explanation:

The sum of the three angles is 90 degrees

x+2x+x+10 = 90

Combine like terms

4x+10 = 90

Subtract 10

4x+10-10 = 90-10

4x= 80

Divide by 4

4x/4=80/4

x = 20

3 0
3 years ago
What is the positive slope of the asymptote of (y+11)^2/100-(x-6)^2/4=1? The positive slope of the asymptote is .
VladimirAG [237]

Answer:

the  positive slope of the asymptote = 5

Step-by-step explanation:

Given that:

\frac{(y+11)^2}{100} -\frac{(x-6)^2}{4} =1

Using the standard form of the equation:

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}= 1

where:

(h,k) are the center of the hyperbola.

and the y term is in front of the x term  indicating that the  hyperbola opens up and down.

a = distance that indicates  how far above and below of the center the vertices of the hyperbola are.

For the above standard equation; the equation for the asymptote is:

y = \pm \frac{a}{b} (x-h)+k

where;

\frac{a}{b} is the slope

From above;

(h,k) = 11, 100

a^2 = 100

a = \sqrt{100}

a = 10

b^2 =4

b = \sqrt{4}

b = 2

y = \pm \frac{10}{2} (x-11)+2

y = \pm 5 (x-11)+2

y = 5x-53 , -5x -57

Since we are to find the positive slope of the asymptote: we have  

\frac{a}{b} to be  the slope in the equation  y = \pm \frac{10}{2} (y-11)+2

\frac{a}{b}  = \frac{10}{2}

\frac{a}{b}  = 5

Thus, the  positive slope of the asymptote = 5

8 0
3 years ago
A team is selling pop corn for a fundraiser. The first week, they sell 500 bags. The second week, they sell 50% more than the fi
Dmitriy789 [7]

Answer:

The percentage decrease from the first week to the third week is 25 %

Step-by-step explanation:

The number of popcorn bags sold on the first week = 500 bags

The second week, they sell 50% more than the first week

So, the number of bags sold on the second week is

= 500 + 50% of 500

=500 +  \frac{50}{100}  \times 500

= 500 + 0.5 \times 500

= 500 + 250

= 750 bags

The third week, they sold 50% less than the second week

So, the number of bags sold on the third week is

=  750 - 50% of 750

= 750 - \frac{50}{100} \times 750

= 750 -0.5 \times 750

= 750 - 375

= 375 bags

Now the percentage decrease from the first week to the third week is

=\frac{500 - 375}{500} \times 100

= \frac{125}{500} \times 100

= 0.25  \times 100

= 25 %

4 0
3 years ago
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