what is area of square

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:

(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 )

Answer:

123456-6-&55674

Step-by-step explanation:

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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:

hi I am a Nepal

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

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:

504

Step-by-step explanation:

Answer:

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

Answer:

Hello,

742/27 (ft)

Step-by-step explanation:

[tex]h_1=14\\\\h_2=\dfrac{14}{3} \\\\h_3=\dfrac{14}{9} \\\\h_4=\dfrac{14}{27} \\\\[/tex]

[tex]d=14+2*\dfrac{14}{3} +2*\dfrac{14}{9} +2*\dfrac{14}{27} \\=14*(1+\dfrac{1}{3}+\dfrac{2}{9} +\dfrac{2}{27} )\\=14*\dfrac{53}{27} \\=\dfrac{742}{27} \\[/tex]

The total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]

What is the total distance?

Distance is a numerical measurement of how far apart objects or points are. It is the actual length of the path travelled from one point to another.

Here given that,

A ball is dropped from a height of [tex]14[/tex] ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.

So, after striking with the ground it covers the distance [tex]14[/tex] ft. so it rebounds to the height is [tex]\frac{1}{3}(14)[/tex].

Then again it hits the ground and covers the distance  [tex]\frac{1}{3}(14)[/tex] and again after rebounding it goes to the height is

[tex]\frac{1(1)}{3(3)}.(14)=\frac{(1)^2}{(3)^2}(14)[/tex]

Then it falls the same distance and goes back to the height

[tex]\frac{1}{3}[/tex] ×[tex](\frac{(1)^2}{(3)^2})[/tex] ×[tex]14[/tex] = [tex]\frac{(1)^3}{(3)^3}(14)[/tex]

So, the total distance travelled is

[tex]14+2[\frac{1}{3}(14)+(\frac{1}{3})^2(14)+(\frac{1}{3})^3(14)+...][/tex]

We take the sum is twice because it goes back to the particular height and falls to the same distance.

[tex]S=14+2(\frac{\frac{1}{3}(14)}{1-\frac{1}{3}})\\\\\\S=\frac{a}{1-r}\\\\\\S=14+2(\frac{\frac{14}{3}}{\frac{2}{3}})\\\\S=14+2(\frac{14}{2})\\\\S=14+2(7)\\\\S=14+14\\\\S=28ft[/tex]

Hence, the total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]

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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:

B

Step-by-step explanation:

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:

(x - 3/8)^2 = x^2 - 3/4x + 9/64

Step-by-step explanation:

Step-by-step explanation:

divide the number with x by 2 and get the square of that number and add that number to this given equation

number with x = -3/4

= x^2 - 3/4 x + 9/64

= (x -3/8) ^2

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

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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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).

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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:

A.x<-8

Step-by-step explanation:

=1/2x<−4

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

= x<-8

Answer:

8,15,22,29

Step-by-step explanation:

the interger a is 7,so to find the next term you have to add 7 plus the 8,

8+7=15

15+7=22

22+7=29

8,15,22,29

I hope this helps

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]

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

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

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

Answer:

B. 240 cm2

Step-by-step explanation:

Chu vi đáy: 10x=40

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

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:

6.986.

Step-by-step explanation:

6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000

We do the multiplications first    ( according to PEMDAS):-

= 6 + 9 * 0.1 + 8 * 0.01 + 6 * 0.001

= 6 + 0.9 + 0.08 + 0006

= 6.9 + 0.086

= 6 986.

The value of the equation in the decimal form is A = 6.986

What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

A = 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000

On simplifying the equation , we get

The value of 6 x 1 = 6

The value of 9 x 1/10 = 0.9

The value of 9 x 1/100 = 0.08

The value of 6 x 1/1000 = 0.006

So , substituting the values in the equation A , we get

A = 6 + 0.9 + 0.08 + 0.006

On simplifying the equation , we get

A = 6.986

Therefore , the value of A is 6.986

Hence , the value of the equation is 6.986

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Answer:

The answer is "Option a, Option b, and Option d".

Step-by-step explanation:

In the given question it is used to stratifying the sampling if the population of this scenario it flights takes off when it is divided via some strata.

Answer:

For each day of the week, randomly select 5% of all flights that depart on that day of the week.

Divide all flights into the following 4 groups on the basis of scheduled departure time: before 9:00 am, 9:00 am to 1:00 pm, 1:00 pm to 5:00 pm, and after 5:00 pm. Randomly select 5% of the flights in each group.

Divide the airports from which the airline's flights depart into 4 regions: Northeast, Northwest, Southwest, and Southeast. Randomly select 5% of all flights departing from airports in each region.

Step-by-step explanation:

ll sampling methods that divide the flights into a small number of mutually exclusive categories are appropriate. These methods include all flights on the basis of a characteristic that might be associated with the variable being investigated and randomly selects a proportionate number of flights from each group.