Find the perimeter of a football field which measures 90m by 60m

Answer:

24 years

Step-by-step explanation:

total ratio =8

older student=40 years

3/8*40 ÷ 5/8=24

Answer:

Answer:I just need points

Step-by-step explanation:

Hey

Answer:

8x-21

----------------------

(2x-7)(2x+7)

Step-by-step explanation:

7                       4

-----------   + ------------

4x^2 -49    2x+7

Factor  ( notice that it is the difference of squares)

7                       4

-----------   + ------------

(2x)^2 - 7^2    2x+7

7                       4

-----------       + ------------

(2x-7)(2x+7)    2x+7

Get a common denominator

7                       4(2x-7)

-----------       + ------------

(2x-7)(2x+7)    (2x-7)(2x+7)

Combine

7 +4(2x-7)

----------------------

(2x-7)(2x+7)  

7 +8x-28

----------------------

(2x-7)(2x+7)  

8x-21

----------------------

(2x-7)(2x+7)  

Answer:

(8x - 21) / (2x + 7)(2x - 7)

Step-by-step explanation:

7 / (4x^2 - 49)+ 4 / (2x + 7)

= 7 / (2x + 7)(2x - 7) + 4 / (2x + 7)

LCM = (2x + 7)(2x - 7)   so we have

(7 + 4(2x - 7) / (2x + 7)(2x - 7)

=   (8x - 21) / (2x + 7)(2x - 7).

Answer:

2

Step-by-step explanation:

the number before the x² is regarded as the identity element

Answer:

The sample proportion of students who prefer vanilla ice cream is 0.56.

Step-by-step explanation:

Sample proportion of students who prefer vanilla ice cream:

Sample of 375 students.

Of those, 210 said they prefer vanilla ice cream.

The proportion is:

[tex]p = \frac{210}{375} = 0.56[/tex]

The sample proportion of students who prefer vanilla ice cream is 0.56.

The set of numbers that Diane can choose is:

{54, 60, 66, 72, 78, 84, 90}

Finding common multiples of 2, 3, and 6:

A number is a multiple of 2 if the number is even.

A number is a multiple of 3 if the sum of its digits is multiples of 3.

A number is a multiple of 6 if it is a multiple of 2 and 3.

Then we only need to look at the first two criteria.

First, let's see all the even numbers in the range (49, 95)

These are:

{50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94}

All of these are multiples of 2.

Now we need to see which ones are multiples of 3.

To do it, we sum its digits and see if that sum is also a multiple of 3.

50: 5 + 0 = 5 this is not multiple of 3.

52: 5 + 2 = 7 this is not multiple of 3.

54: 5 + 4 = 9 this is multiple of 3, so 54 is a possible number.

And so on, we will find that the ones that are multiples of 3 are:

54: 5 + 4 = 9.

60: 6 + 0 = 6

66: 6 + 6 = 12

72: 7 + 2 = 9

78: 7 + 8 = 15

84: 8 + 4 = 12

90:9 + 0 = 9

Then the numbers that Diane could choose are:

{54, 60, 66, 72, 78, 84, 90}

If you want to learn more about multiples, you can read:

https://brainly.com/question/1553674

Answer:

16. f(y-2) = 2(y-2)²-5

= 2(y²-4y+4)-5

= 2y²-8y+8-5

= 2y²-8y+3

17. f(a+h)-f(a) = 2(a+h)²-5-(2a²-5)

= 2(a²+2ah+h²)-5-2a²+5

= 2a²+4ah+h²-2a²

= h²+4ah

Answer: 7 / 50

Step-by-step explanation:

Given

0.14

Convert to 100-denominator fraction

= 14 ÷ 100

= 14/100

Divide both numerator and denominator by 2

=(14 ÷ 2) / (100 ÷ 2)

=7 / 50

Hope this helps!! :)

Please let me know if you have any questions

Answer:

False

Step-by-step explanation:

To find the inverse of a function, switch the variables and solve for y.

The inverse of f(n)=-(n+1)^3:

[tex]y=-(n+1)^3[/tex]

[tex]n=-(y+1)^3[/tex]

[tex]\sqrt[3]{n} =-(y+1)[/tex]

[tex]\sqrt[3]{n} =-y-1[/tex]

[tex]\sqrt[3]{n} +1=-y[/tex][tex]-(\sqrt[3]{n} +1)=y[/tex]

[tex]-\sqrt[3]{n} -1=y[/tex]

Answer:

False

Step-by-step explanation:

12321-3x3x37x37

(3)^2×(37)^2

square root = 3×37=111

Hope it helps you..!!

Answer:

answer 11

Step-by-step explanation:

I think it the right answer

Answer:

53.68

Step-by-step explanation:

tan54 = bc/39

bc = 39tan54

Step-by-step explanation:

Hey there!

From the given figure;

Angle CAB = 54°

Side AC = 39

To find: side BC

Taking Angle CAB as reference angle;

Perpendicular (p) = BC = ?

Base (b) = AC = 39

Hypotenuse (h) = AB

Taking the ratio of tan;

[tex] \tan( \alpha ) = \frac{p}{b} [/tex]

Keep value;

[tex] \tan(54) = \frac{bc}{39} [/tex]

Simplify it;

1.376381*39 = BC

Therefore, BC = 53.678.

Hope it helps!

Answer:

Does the answer help you?

Answer:

Pencils = 325 ; Pens = 975 ; Markers = 650

Step-by-step explanation:

Let :

Number of Pencils = x

Number of pens = y

Number of markers = z

2 times as many markers as pencils

z = 2x

3 times as many pens as pencils

y = 3x

x + y + z = 1950

Write z and y in terms of x in the equation :

x + 3x + 2x = 1950

6x = 1950

Divide both sides by 6

6x / 6 = 1950 / 6

x = 325

Number of pencils = 325

Pens = 3 * 325 = 975

Markers = 2 * 325 = 650

Pencils = 325 ; Pens = 975 ; Markers = 650

Answer:

The polynomial is:

[tex]p(x) = -x^3 - 2x^2 + 5x + 6[/tex]

Step-by-step explanation:

Zeros of a function:

Given a polynomial f(x), this polynomial has roots [tex]x_{1}, x_{2}, x_{n}[/tex] such that it can be written as: [tex]a(x - x_{1})*(x - x_{2})*...*(x-x_n)[/tex], in which a is the leading coefficient.

Zeros of −3, −1, and 2

This means that [tex]x_1 = -3, x_2 = -1, x_3 = 2[/tex]. Thus

[tex]p(x) = a(x - x_{1})*(x - x_{2})*(x-x_3)[/tex]

[tex]p(x) = a(x - (-3))*(x - (-1))*(x-2)[/tex]

[tex]p(x) = a(x+3)(x+1)(x-2)[/tex]

[tex]p(x) = a(x^2+4x+3)(x-2)[/tex]

[tex]p(x) = a(x^3+2x^2-5x-6)[/tex]

Passes through the point (1,12).

This means that when [tex]x = 1, p(x) = 12[/tex]. We use this to find a.

[tex]12 = a(1 + 2 - 5 - 6)[/tex]

[tex]-12a = 12[/tex]

[tex]a = -\frac{12}{12}[/tex]

[tex]a = -1[/tex]

Thus

[tex]p(x) = -(x^3+2x^2-5x-6)[/tex]

[tex]p(x) = -x^3 - 2x^2 + 5x + 6[/tex]

Assuming the car's speed [tex]\frac{630}{14}=45\mathrm{mph}[/tex] does not change, the car will travel [tex]45\cdot42=\boxed{1890}[/tex] miles.

Hope this helps :)

Explanation:

Let [tex]f(x) = \sqrt{25x}[/tex] and [tex]g(x) = \frac{x^2}{25}[/tex]. The differential volume dV of the cylindrical shells is given by

[tex]dV = 2\pi x[f(x) - g(x)]dx[/tex]

Integrating this expression, we get

[tex]\displaystyle V = 2\pi\int{x[f(x) - g(x)]}dx[/tex]

To determine the limits of integration, we equate the two functions to find their solutions and thus the limits:

[tex]\sqrt{25x} = \dfrac{x^2}{25}[/tex]

We can clearly see that x = 0 is one of the solutions. For the other solution/limit, let's solve for x by first taking the square of the equation above:

[tex]25x = \dfrac{x^4}{(25)^2} \Rightarrow \dfrac{x^3}{(25)^3} = 1[/tex]

or

[tex]x^3 =(25)^3 \Rightarrow x = \pm25[/tex]

Since we are rotating the functions around the y-axis, we are going to use the x = 25 solution as one of the limits. So the expression for the volume of revolution around the y-axis is

[tex]\displaystyle V = 2\pi\int_0^{25}{x\left(\sqrt{25x} - \frac{x^2}{25}\right)}dx[/tex]

[tex]\displaystyle\:\:\:\:=10\pi\int_0^{25}{x^{3/2}}dx - \frac{2\pi}{25}\int_0^{25}{x^3}dx[/tex]

[tex]\:\:\:\:=\left(4\pi x^{5/2} - \dfrac{\pi}{50}x^4\right)_0^{25}[/tex]

[tex]\:\:\:\:=4\pi(3125) - \pi(7812.5) = 14726.2[/tex]

Answer:

Step-by-step explanation:

In order to find out how long it takes for the rocket to hit the ground, we only need set that position equation equal to 0 (that's how high something is off the ground when it is sitting ON the ground) and factor to solve for t:

[tex]0=-4.9t^2+112t+395[/tex]

Factor that however you are factoring in class to get

t = -3.1 seconds and t = 25.9 seconds.

Since time can NEVER be negative, it takes the rocket approximately 26 seconds to hit the ground.

Answer:

Step-by-step explanation:

Answer:

f(x) can not be equal to g(x)

Step-by-step explanation:

If the result is possible:

f(x) = g(x)

x + 8 = x + 2

x + 8 - (x + 2) = x + 2 - (x + 2)

6 = 0

Because 6 can't be equal to 0, so do f(x) can't be equal to g(x)

Answer:

45°

Step-by-step explanation:

[tex] \sin \: m\angle F = \frac{EG}{FG} \\ \\ \sin \: m\angle F = \frac{2 \sqrt{11} }{2 \sqrt{22} } \\ \\ \sin \: m\angle F = \frac{\sqrt{11} }{ \sqrt{22} } \\ \\ \sin \: m\angle F = \frac{1}{ \sqrt{2} } \\ \\ \sin \: m\angle F = \sin \: 45 \degree \\ \\ \huge \boxed{ \purple{m\angle F = 45 \degree }}[/tex]

Answer:

606.6 and 233.3 respectively

Step-by-step explanation:

Let the speed of plane in still air be x and the speed of wind be y.

ATQ, (x+y)*9=7560 and (x-y)*9=3360. Solving it, we get x=606.6 and y=233.3

Answer:

1a 300

b 180

c 330

d 1470

e 180

f 7344

Step-by-step explanation:

Answer:Mark Brainliest please

Answer is 4.86 which is rounded to 5

Step-by-step explanation:

Cos 40 degree = VW/7

0.694 =VW/7

0.694 * 7 =VW

4.858 =VW

VW=4.86 is the answer

Hi there!

[tex]\large\boxed{a + b + c = 6}[/tex]

We can begin by using the point (0, 1).

At the graph's y-intercept, where x = 0, y = 1, so:

1 = a(0)² + b(0) + c

c = 1

We can now utilize the first point given (-3, 10):

10 = a(-3)² + b(-3) + 1

Simplify:

9 = 9a - 3b

Divide all terms by 3:

3 = 3a - b

Rearrange to solve for a variable:

b = 3a - 3

Now, use the other point:

15 = a(2)² + 2(3a - 3) + 1

14 = 4a + 6a - 6

Solve:

20 = 10a

2 = a

Plug this in to solve for b:

b = 3a - 3

b = 3(2) - 3 = 3

Add all solved variables together:

2 + 3 + 1 = 6

Answer:

See below

Step-by-step explanation:

we want to prove that A.M, G.M. and H.M between any two unequal positive numbers satisfy the following relations.

well, to do so let the two unequal positive numbers be [tex]\text{$x_1$ and $x_2$}[/tex] where:

the AM,GM and HM of [tex]x_1[/tex] and[tex] x_2[/tex] is given by the following table:

[tex]\begin{array}{ |c |c|c | } \hline AM& GM& HM\\ \hline \dfrac{x_{1} + x_{2}}{2} & \sqrt{x_{1} x_{2}} & \dfrac{2}{ \frac{1}{x_{1} } + \frac{1}{x_{2}} } \\ \hline\end{array}[/tex]

[tex] \displaystyle \rm AM \times HM = \frac{x_{1} + x_{2}}{2} \times \frac{2}{ \frac{1}{x_{1} } + \frac{1}{x_{2}} } [/tex]

simplify addition:

[tex] \displaystyle \frac{x_{1} + x_{2}}{2} \times \frac{2}{ \dfrac{x_{1} + x_{2}}{x_{1} x_{2}} } [/tex]

reduce fraction:

[tex] \displaystyle x_{1} + x_{2} \times \frac{1}{ \dfrac{x_{1} + x_{2}}{x_{1} x_{2}} } [/tex]

simplify complex fraction:

[tex] \displaystyle x_{1} + x_{2} \times \frac{x_{1} x_{2}}{x_{1} + x_{2}} [/tex]

reduce fraction:

[tex] \displaystyle x_{1} x_{2}[/tex]

rewrite:

[tex] \displaystyle (\sqrt{x_{1} x_{2}} {)}^{2} [/tex]

[tex] \displaystyle AM \times HM = (GM{)}^{2} [/tex]

hence, PROVEN

[tex] \displaystyle x_{1} > x_{2}[/tex]

square root both sides:

[tex] \displaystyle \sqrt{x_{1} }> \sqrt{ x_{2}}[/tex]

isolate right hand side expression to left hand side and change its sign:

[tex]\displaystyle\sqrt{x_{1} } - \sqrt{ x_{2}} > 0[/tex]

square both sides:

[tex]\displaystyle(\sqrt{x_{1} } - \sqrt{ x_{2}} {)}^{2} > 0[/tex]

expand using (a-b)²=a²-2ab+b²:

[tex]\displaystyle x_{1} -2\sqrt{x_{1} }\sqrt{ x_{2}} + x_{2} > 0[/tex]

move -2√x_1√x_2 to right hand side and change its sign:

[tex]\displaystyle x_{1} + x_{2} > 2 \sqrt{x_{1} } \sqrt{ x_{2}}[/tex]

divide both sides by 2:

[tex]\displaystyle \frac{x_{1} + x_{2}}{2} > \sqrt{x_{1} x_{2}}[/tex]

[tex]\displaystyle \boxed{ AM>GM}[/tex]

again,

[tex]\displaystyle \bigg( \frac{1}{\sqrt{x_{1} }} - \frac{1}{\sqrt{ x_{2}}} { \bigg)}^{2} > 0[/tex]

expand:

[tex]\displaystyle \frac{1}{x_{1}} - \frac{2}{\sqrt{x_{1} x_{2}} } + \frac{1}{x_{2} }> 0[/tex]

move the middle expression to right hand side and change its sign:

[tex]\displaystyle \frac{1}{x_{1}} + \frac{1}{x_{2} }> \frac{2}{\sqrt{x_{1} x_{2}} }[/tex]

[tex]\displaystyle \frac{\frac{1}{x_{1}} + \frac{1}{x_{2} }}{2}> \frac{1}{\sqrt{x_{1} x_{2}} }[/tex]

[tex]\displaystyle \rm \frac{1}{ HM} > \frac{1}{GM} [/tex]

cross multiplication:

[tex]\displaystyle \rm \boxed{ GM >HM}[/tex]

hence,

[tex]\displaystyle \rm A.M>G.M>H.M[/tex]

PROVEN

Answer:

Step-by-step explanation:

9514 1404 393

Answer:

  (x, y) = (4, -4)

Step-by-step explanation:

A graphing calculator makes graphing very easy. The attachment shows the solution to be (x, y) = (4, -4).

__

The equations are in slope-intercept form, so it is convenient to start from the y-intercept and use the slope (rise/run) to find additional points on the line.

The first line can be drawn by staring at (0, -2) and moving down 1 grid unit for each 2 to the right.

The second line can be drawn by starting at (0, 2) and moving down 3 grid units for each 2 to the right.

The point of intersection of the lines, (4, -4), is the solution to the system of equations.