Evaluate 3|x|+2x-1 when x = -5.

B.The failures of the past should inspire people to accomplish more in the future.

Answer:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]V(50) = 2548.17[/tex]        [tex]V(100) = 10098.10[/tex]       [tex]V(1000) = 999201.78[/tex]

[tex]x = 54.78[/tex]

Step-by-step explanation:

Given

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

[tex]C_1(x) = \frac{x}{x+1}[/tex]

[tex]C_1(x) = \frac{2}{x-3}[/tex]

[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]

Solving (a): Expression for V(x)

We have:

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

Substitute known values

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solving (b): Simplify V(x)

We have:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solve the expression in bracket

[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

Factor out x

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]

Express as difference of two squares

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]

Cancel out x - 3

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Solving (c): V(50), V(100), V(1000)

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Substitute 50 for x

[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]

[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]

[tex]V(50) = 2548.17[/tex]

Substitute 100 for x

[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]

[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]

[tex]V(100) = 10098.10[/tex]

Substitute 1000 for x

[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]

[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]

[tex]V(1000) = 999201.78[/tex]

Solving (d): V(x) = 3000, find x

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Cross multiply

[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]

Equate to 0

[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]

Open brackets

[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]

Collect like terms

[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]

[tex]x^3 + x^2 -3001x -2994 = 0[/tex]

Solve using graphs (see attachment)

[tex]x = -54.783[/tex] or

[tex]x = -0.998[/tex] or

[tex]x = 54.78[/tex]

x can't be negative. So:

[tex]x = 54.78[/tex]

Answer:

Step-by-step explanation:

Here's how you convert:

[tex]\sqrt[n]{x^m}=x^{\frac{m}{n}[/tex]  The little number outside the radical, called the index, serves as the denominator in the rational power, and the power on the x inside the radical serves as the numerator in the rational power on the x.

A couple of examples:

[tex]\sqrt[3]{x^4}=x^{\frac{4}{3}[/tex]

[tex]\sqrt[5]{x^7}=x^{\frac{7}{5}[/tex]

It's that simple. For your problem in particular:

[tex]\sqrt[3]{7}[/tex] is the exact same thing as [tex]\sqrt[3]{7^1}=7^{\frac{1}{3}[/tex]

Answer:

A.x<-8

Step-by-step explanation:

=1/2x<−4

=2*(1/2x)< (2)*(-4)

= x<-8

Answer:

B. 240 cm2

Step-by-step explanation:

Chu vi đáy: 10x=40

Diện tích xung quanh:  Sxq=1/2 x40x12=240

Answer:

79/9  or 8 7/9

Step-by-step explanation:

1+9x=80

1 +9x = 90

Subtract 1 from each side

1+9x-1 = 80-1

9x = 79

divide by 9

9x/9 = 79/9

x = 8 7/9

Answer:

Step-by-step explanation:

1+9x = 80

9x = 80-1

9x = 79

x = 79/9

or

x = 8 7/9

or

x ≈ 8.78 (x = about 8.78) rounded

Answer:

123456-6-&55674

Step-by-step explanation:

rdcfvvzxv.

dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see

recall see

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

Website 1 : 10357, 10537, 10767, 10561, 10544, 10581, 10602, 10665, 10335, 10419, 10737, 10410, 10485, 10601, 10458, 10472, 10435, 10375, 10436, 10510, 10345, 10559, 10520, 10425, 10351, 10465, 10491, 10671, 10366, 10440, 10618, 10606, 10406, 10538, 10449, 10462

Mean, xbar = ΣX/ n ; n = sample size = 36

Xbar = 377999 / 36 = 10499.9722

Standard deviation, s = √[(x - xbar)² / (n-1]

Using calculator :

Standard deviation (Website 1 :), s = 110.239865

Website 2 : 11067, 11029, 10888, 10789, 10914, 10663, 10787, 11140, 11042, 11074, 10868, 10853, 10900, 11088, 10991, 10928, 10959, 11126, 11033, 11114, 11150, 11155, 11027, 10900, 11015, 11123, 10953, 11181, 10855, 10731, 10971, 10770, 11070, 11122, 11018, 10903

Mean, xbar = ΣX/ n ; n = sample size = 36

Xbar = 395197 / 36 = 10977.6944

Standard deviation, s = √[(x - xbar)² / (n-1]

Using calculator :

Standard deviation (Website 2), s = 132.617995

2.)

Yes, the viewership between the two websites are different with the second website has a higher mean viewership with a mean of 10977.6944.

3.)

The probability of 12000 views per month on each website :

Probability = Mean viewership per month / required viewership

Website 1 :

P(12000) = 10499.9722 / 12000 = 0.8749

Website 2 :

P(12000) = 10977.6944 / 12000 = 0.9148

4.)

More consistent website :

We use the standard deviation value, the higher the standard deviation, the higher the variability :

Website 1 should be more consistent has it has a Lower standard deviation score, hence, should show lower variability than website 2.

5.)

Website suitable for advertisement should be one with higher viewership per month in other to reach a larger audience. Hence, website 2 should be recommended for advertisement.

Answer:

hi I am a Nepal

[tex] {233333}^{2332} [/tex]

Answer:

V = 1071.79 yd^3

Step-by-step explanation:

The volume of a cone is

V = 1/3 pi r^2 h  where r is the radius and h is the height

We are given a diameter of 16 so the radius is 1/2 of the diameter or 8

The height is 16

V = 1/3 ( 3.14) (8)^2 ( 16)

V = 1071.78666 yd^3

Rounding to the nearest hundredth

V = 1071.79 yd^3

[tex] \large\begin{gathered} {\underline{\boxed{ \rm {\red{Volume \: of \: Cone \: = \: \pi \: {r}^{2} \: \frac{h}{3} }}}}}\end{gathered}[/tex]

[tex] \bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{Diameter}{2} \\ [/tex]

[tex]\bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{16 \: yd}{2} \\ [/tex]

[tex]\bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{ \cancel{16 \: yd} \: \: ^{8} }{ \cancel{2}} \\ [/tex]

[tex]\bf{\red{ \longrightarrow}} \tt \: \large{\bf{{{\color{navy}{r \: = \: 8 \: yd}}}}}[/tex]

[tex]\bf{\red{ \longrightarrow}} \tt \: \: \large{\bf{{{\color{navy}{h \: = \: 16 \: yd}}}}}[/tex]

[tex] \bf \large \longrightarrow \: \: 3.14 \: \times \: {8}^{2} \: \times \: \frac{16}{3} \\ [/tex]

[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: \frac{16}{3} \\ [/tex]

[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: \frac{ \cancel{16} \: \: ^{5.33} }{ \cancel{3}} \\ [/tex]

[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: 5.33[/tex]

[tex]\bf \large \longrightarrow \: \:200.96 \: \times \: 5.3[/tex]

[tex]\bf \large \longrightarrow \: \:1071.79[/tex]

Option (A) is the correct answer

Answer:

a x^2/2 is a polynomial because the power of x is 2 which is a positive whole number but 2/x^2 is not a polynomial because the power of x is -2 which is negative whole number.

b.in

[tex] \sqrt{2 x} [/tex]

the power if x will be

[tex]x {}^{ \frac{1}{2} } [/tex]

which is not a whole number so it is not a polynomial.

but in

[tex] \sqrt{2} x[/tex]

the power if x is a positive whole number.so it is a polynomial.

c.the greatest power of variable of the term is called degree of polynomial

The equation of the line through (0, 1) and (c, 0) is

y - 0 = (0 - 1)/(c - 0) (x - c)   ==>   y = 1 - x/c

Let L denote the given lamina,

L = {(x, y) : 0 ≤ x ≤ c and 0 ≤ y ≤ 1 - x/c}

Then the center of mass of L is the point [tex](\bar x,\bar y)[/tex] with coordinates given by

[tex]\bar x = \dfrac{M_x}m \text{ and } \bar y = \dfrac{M_y}m[/tex]

where [tex]M_x[/tex] is the first moment of L about the x-axis, [tex]M_y[/tex] is the first moment about the y-axis, and m is the mass of L. We only care about the y-coordinate, of course.

Let ρ be the mass density of L. Then L has a mass of

[tex]\displaystyle m = \iint_L \rho \,\mathrm dA = \rho\int_0^c\int_0^{1-\frac xc}\mathrm dy\,\mathrm dx = \frac{\rho c}2[/tex]

Now we compute the first moment about the y-axis:

[tex]\displaystyle M_y = \iint_L x\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}x\,\mathrm dy\,\mathrm dx = \frac{\rho c^2}6[/tex]

Then

[tex]\bar y = \dfrac{M_y}m = \dfrac{\dfrac{\rho c^2}6}{\dfrac{\rho c}2} = \dfrac c3[/tex]

but this clearly isn't independent of c ...

Maybe the x-coordinate was intended? Because we would have had

[tex]\displaystyle M_x = \iint_L y\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}y\,\mathrm dy\,\mathrm dx = \frac{\rho c}6[/tex]

and we get

[tex]\bar x = \dfrac{M_x}m = \dfrac{\dfrac{\rho c}6}{\dfrac{\rho c}2} = \dfrac13[/tex]

The center of mass for a uniform triangular shape is on its centroid. The y-coordinate of the center of mass of the lamina is 1/3 (independent of c).

If the surface is plane triangle approximately and mass is uniformally distributed, then its center of mass will lie on the centroid of that triangle.

The point of intersection of a triangle's medians is its centroid (the lines joining each vertex with the midpoint of the opposite side).

If the triangle has its vertices as  [tex](x_1, y_1), (x_2, y_2) , \: (x_3, y_3)[/tex], then the coordinates of the centroid of that triangle is given by:

[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)[/tex]

For this case, the  triangular lamina has vertices (0, 0), (0, 1) and (c, 0)

Assuming its mass is spread regularly, the coordinates of its center of mass would be:

[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)\\\\(x,y) = \left( \dfrac{0+0+c}{3} + \dfrac{0+1+0}{3} \right) = (c/3, 1/3)[/tex]

Thus, the y-coordinate of the center of mass of the lamina is 1/3 (independent of c).

Learn more about centroid here:

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9514 1404 393

Answer:

  525 km

Step-by-step explanation:

Let d represent the distance to the town. Let s represent the nominal speed of the car. The relation between time, speed, and distance is d = st.

  t1 = d/s

  t2 = d/(s+15)

  t1 : t2 = 6 : 5 . . . increasing the speed reduces the time

Substituting for t1 and t2, we have ...

  (d/s)/(d/(s+15)) = 6/5

  (s +15)/s = 6/5

  1 +15/s = 1 +1/5

  s = 5·15 = 75 . . . . nominal speed in km/h

__

Decreasing the speed increases the time.

  d/75 +(105/60) = d/(75-15)

  d(60/75) +105 = d . . . . . . multiply by 60

  105 = d/5 . . . . . . . . . . . subtract 4/5d

  525 = d . . . . . . . . . . multiply by 5

The distance traveled by the car is 525 km.  

Answer:

[tex] = 2 {}^{2} - 3(2) = - 2 \\ 3 {}^{2} - 3(3) = 0 \\ 4 {}^{2} - 3(4) = 4 \\ 5 {}^{2} - 3(5) = 10[/tex]

Answer:

504

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Answer:

93 child tickets

Step-by-step explanation:

Create a system of equations where c is the number of child tickets sold and a is the number of adult tickets sold:

c + a = 171

5.80c + 9.70a = 1296.00

Solve by elimination by multiplying the top equation by -9.7:

-9.7c - 9.7a = -1658.7

5.8c + 9.7a = 1296

Add these together and solve for c:

-3.9c = -362.7

c = 93

So, 93 child tickets were sold.

Answer:

ACB = JCB

Step-by-step explanation:

ASA means angle - (included) side - angle.

we have one angle confirmed.

we have the connected side BC confirmed (the diagram shows that the side is shared, so it is not only congruent, it is actually identical).

now we need confirmation for the angle at the other end point of that side.

Answer:

D

Step-by-step explanation:

sqrt_{4}(81)^5=(81^(5))^(1/4)=81^(5/4)

Answer:

D

Step-by-step explanation:

if it was properly typed, it would have been All of the above but the most correct option is D.

Answer:

Ann has $30

Ryan has $63

Keith has $21

Step-by-step explanation:

First I assigned variables to the amounts each owns. It doesn't really matter what letters, but I said x represents Ann's money, y represents Ryan's money, and z represents Keith's money.

We know that with all their money added up, they have 114. This gives us our first equation;

114=x+y+z

Ryan has 3 times what Keith has, which, in terms of variables, is

y=3z

Keith has nine dollars less than Ann. Less, in word problems, always means subtraction. This gives us our third equation:

z=x-9

Okay so the impulse is to put z=x-9 and y=3z into the first equation, but the goal is to create an equation with only one variable so this will not work.

What we do instead is sub z=x-9 into y=3z. This gives

y=3(x-9)=3x-27

Now that both y and z are in terms of x, we plug those equations into the first and then solve for x;

114=x+3x-27+x-9

114=5x-36

150=5x

30=x

Now that we know x, we can find z and y using the equations from earlier:

z=30-9=21

y=3(21)=63

x=30, y=63, and z=21, thus Ann has $30, Ryan has $63, and Keith has $21.

Answer:

(137.78, 47.72)

Step-by-step explanation:

(I just finished this assignment.)

Tre's position at the pitcher's mound as the point (42.78, 42.78).

                                                                                  (     x    ,    y     )

Hector is about 95 feet away from the pitcher's mound horizontal, (x axis).

Since we already have the correct y-coordinate, we need to solve for the correct x-coordinate.

  x = 95 + 42.78

         ↓ ↓ ↓

95 + 42.78 = 137.72

Now all you need to do is write out the coordinates.

Hector's coordinates are (137.72, 47.78 )

Step-by-step explanation:

41a+11a-2b+3c

52a-2b+3c

Step-by-step explanation:

(41a-2b+3c) +11a

open the bracket

41a-2b+3c+11a

collect like terms

41a+11a-2b+3c

52a-5bc

so final answer is 52a-5bc

because they are not like terms they are unlike terms

Answer:

Step-by-step explanation:

The combinations are:

Total number of combinations:

Correct choice is B

there are 8different combinations are modeled by the diagram.

Answer:

Solution given:

orange:2

grape:2

apple:2

grapefruit:2

no of term:4

now

total no. of combination ia 4*2=8

Hence the time that the ball will be height than 12 feet off the ground is 4secs

Given the expression for calculating the height in  feet as;

h(t) = -4t²+16t

If the ball is higher than 12feet, h(t) > 12

Substituting h = 12 into the expression

-4t²+16t > 12

-4t²+16t - 12 > 0

4t²- 16t + 12 > 0

t²- 4t + 3 > 0

Factorize

(t²- 3t)-(t + 3) > 0

t(t-3)-1(t-3) > 0

(t-1)(t-3)>0

t > 1 and 3secs

Hence the time that the ball will be height than 12 feet off the ground is 4secs

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Answer:

Beginning (t=0) population with flu is 23.

After 4 weeks, population with flu is 1250.

After an infinite amount of weeks, the population witf flu is 102000

Step-by-step explanation:

First question asks you to replace t with 0 because it says beginning.

102000/(1+4400e^-0)=102000/(1+4400)=102000/4401=23.17655 approximately. To nearest whole number this is 23.

After 4 weeks means we replace t with 4:

102000/(1+4400e^-4)

Calculator time:

1250.17142 which to nearest whole number is 1250

If t is super large, then e^-t is super close to 0.

So the limiting number is

102000/(1+4400×0)=102000/1=102000

Answer:

Option B, 1

Step-by-step explanation:

tan 45° = 1/1 = 1

The line PQ of the triangle is an altitude. The correct option is A.

A line segment passing through a triangle's vertex and running perpendicular to the line containing the base is the triangle's height in geometry.

The extended base of the altitude is the name given to this line that contains the opposing side. The foot of the altitude is the point at where, the extended base and the height converge.

In the given triangle the line segment PQ is passing through a triangle's vertex and running perpendicular to the line containing the base is the triangle's height in geometry.

Therefore, the line PQ of the triangle is an altitude. The correct option is A.

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