Which expression represents the probability of choosing an item from the intersection of A and B

Answer:

B

Step-by-step explanation:

1) initial circumference

radius * 2 * pi = 24 * pi = 75,398224

2) final circumference

diameter * pi = 48 * pi = 150,796447

150,796447 - 75,398224 = 75.36 m

220 cubic units.

Answer:

Solution given:

for small cylinder

r=1

and for large cylinder

R=5+1=6

height for both [h]=2

Now

Volume of solid=πR²h-πr²h=πh(R²-r²)

=3.14*2(6²-1²)=219.8 =220 units ³.

Small cylinder is r=1

Large cylinder is R= 5+1 =6

Height (h) =2

Volume of solid,

→ πR²h-πr²h

→ πh(R²-r²)

→ 3.14 × 2(6²-1²)

→ 219.8

→ 220 cubic units

Answer:

It would be the letter B :)

Answer:

4p² - 9p

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Distributive Property

Algebra I

Step-by-step explanation:

Step 1: Define

Identify

3p(2p - 9) - 2p(-9 + p)

Step 2: Simplify

Answer:

answer is no. a -1/8 for sure

Answer:

[tex]a_1 = 2[/tex]

[tex]a_2 = 3[/tex]

[tex]a_3 = 5[/tex]

[tex]a_4 = 8[/tex]

[tex]a_5 = 13[/tex]

Step-by-step explanation:

Given

[tex]a_n = a_{n-1} + a_{n-2}[/tex]

[tex]a_1 = 2[/tex]

[tex]a_2 = 3[/tex]

Solving (a): The first term

This has already been given as:

[tex]a_1 = 2[/tex]

Solving (b): The second term

This has already been given as:

[tex]a_2 = 3[/tex]

Solving (c): The third term

This is calculated as:

[tex]a_n = a_{n-1} + a_{n-2}[/tex]

[tex]a_3 = a_{3-1} +a_{3-2}[/tex]

[tex]a_3 = a_2 +a_1[/tex]

[tex]a_3 = 3 +2[/tex]

[tex]a_3 = 5[/tex]

Solving (d): The fourth term

This is calculated as:

[tex]a_n = a_{n-1} + a_{n-2}[/tex]

[tex]a_4 = a_{4-1} +a_{4-2}[/tex]

[tex]a_4 = a_3 +a_2[/tex]

[tex]a_4 = 5+3[/tex]

[tex]a_4 = 8[/tex]

Solving (e): The fifth term

This is calculated as:

[tex]a_n = a_{n-1} + a_{n-2}[/tex]

[tex]a_5 = a_{5-1} +a_{5-2}[/tex]

[tex]a_5 = a_4 +a_3[/tex]

[tex]a_5 = 8+5[/tex]

[tex]a_5 = 13[/tex]

Given:

The function is

[tex]f(x)=(x-5)^2(3-x)^2[/tex]

To find:

The derivative of the given function.

Solution:

Chain rule of differentiation:

[tex][f(g(x))]'=f'(g(x))g'(x)[/tex]

Product rule of differentiation:

[tex][f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)[/tex]

We have,

[tex]f(x)=(x-5)^2(3-x)^2[/tex]

Differentiate with respect to x.

[tex]f'(x)=(x-5)^2\dfrac{d}{dx}(3-x)^2+(3-x)^2\dfrac{d}{dx}(x-5)^2[/tex]

[tex]f'(x)=(x-5)^2[2(3-x)(0-1)]+(3-x)^2[2(x-5)(1-0)][/tex]

[tex]f'(x)=(x^2-10x+25)(-6+2x)+(9-6x+x^2)(2x-10)[/tex]

[tex]f'(x)=(x^2)(-6)+(-10x)(-6)+(25)(-6)+(x^2)(2x)-10x(2x)+25(2x)+(9)(2x)+(-6x)(2x)+x^2(2x)+9(-10)+(-6x)(-10)+x^2(-10)[/tex]

On further simplification, we get

[tex]f'(x)=-6x^2+60x-150+2x^3-20x^2+50x+18x-12x^2+2x^3-90+60x-10x^2[/tex]

[tex]f'(x)=(2x^3+2x^3)+(-6x^2-20x^2-12x^2-10x^2)+(60x+50x+18x+60x)+(-90-150)[/tex]

[tex]f'(x)=4x^3-48x^2+188x-240[/tex]

Therefore, the derivative of the given function is [tex]f'(x)=4x^3-48x^2+188x-240[/tex].

Answer:

[tex]A_1 = 12[/tex]

[tex]A_n = A_{n-1} + 3[/tex]

Step-by-step explanation:

Given

[tex]A_n =12+(n-1)3[/tex]

Required

Write as recursive

We have:

[tex]A_n =12+(n-1)3[/tex]

Open bracket

[tex]A_n =12+3n-3[/tex]

[tex]A_n =12-3+3n[/tex]

[tex]A_n =9+3n[/tex]

Calculate few terms

[tex]A_1 =9+3*1 = 9 + 3 = 12[/tex]

[tex]A_2 =9+3*2 = 9 + 6 = 15[/tex]

[tex]A_3 =9+3*3 = 9 + 9 = 18[/tex]

The above shows that the rule is to add 3.

So, we have:

[tex]A_1 = 12[/tex]

[tex]A_n = A_{n-1} + 3[/tex]

Answer:

= -9pq

Step-by-step explanation:

=8pq + (-17pq)

=8pq-17pq

= -9pq

Answer:

Step-by-step explanation:

P=2(w)+2(l)

34=2(6.5)+2(l)

34=13+2(l)

21=2(l)

10.5=l

Answer:

[tex](a)\ 6m(3m + 21) = 9(2m^2 + 14m)[/tex]

(b) Yes, it is divisible by 9

Step-by-step explanation:

Given

[tex]6m(3m + 21)[/tex]

Solving (a): Complete the blanks

[tex]6m(3m + 21) = 9 [\ ][/tex]

Expand the bracket

[tex]6m(3[m + 7]) = 9 [\ ][/tex]

[tex]6m*3(m + 7) = 9 [\ ][/tex]

Express 6m as 2m * 3

[tex]2m*3*3(m + 7) = 9 [\ ][/tex]

[tex]2m*9(m + 7) = 9 [\ ][/tex]

Rewrite as:

[tex]9 * 2m(m + 7) = 9 [\ ][/tex]

Multiply the bracket by 2m

[tex]9 * (2m^2 + 14m) = 9 [\ ][/tex]

Divide both sides by 9

[tex]2m^2 + 14m = [\ ][/tex]

Hence, the bracket will be filled with: [tex]2m^2 + 14m[/tex]

So:

[tex]6m(3m + 21) = 9(2m^2 + 14m)[/tex]

Solving (b): Is [tex]6m(3m + 21)[/tex] divisible by 9?

In (a), we have:

[tex]6m(3m + 21) = 9(2m^2 + 14m)[/tex]

The leading factor "9" implies that the expression is divisible by 9

Let n represent the interger l, the three consective intergers are represented by

[tex]n[/tex]

[tex]n + 2[/tex]

[tex]n + 4[/tex]

The second one represent

[tex](n + 2) {}^{2} + 76 =( n + 4) {}^{2} [/tex]

Simplify both sides

[tex]n {}^{2} + 4n + 4 + 76 = {n}^{2} + 8n + 16[/tex]

[tex] {n}^{2} + 4n + 4 = {n}^{2} + 8n - 60[/tex]

[tex]4n + 4 = 8n - 60[/tex]

[tex]4n + 64= 8n[/tex]

[tex]64= 4n[/tex]

[tex]n = 16[/tex]

The intergers are 16,18,20

9514 1404 393

Answer:

 1.60

Step-by-step explanation:

The growth factor is 1 more than the growth rate:

  1 + 60% = 1 + 0.60 = 1.60 = growth factor

Answer:

Step-by-step explanation:

Answer:

The work done is 202.50Nm

Step-by-step explanation:

Given

[tex]F =450N[/tex]

[tex]x_1 = 30cm[/tex]

[tex]x_2 = 60cm[/tex]

Required

The work done

First, we calculate the spring constant (k)

[tex]F = kx_1[/tex]

[tex]450N = k *30cm[/tex]

[tex]k = \frac{450N}{30cm}[/tex]

[tex]k =15N/cm[/tex]

So:

[tex]F = kx_1[/tex]

[tex]F(x) = 15x[/tex]

The work done using Hooke's law is:

[tex]W =\int\limits^a_b {F(x)} \, dx[/tex]

This gives:

[tex]W =\int\limits^{60}_{30} {15x} \, dx[/tex]

Rewrite as:

[tex]W =15\int\limits^{60}_{30} {x} \, dx[/tex]

Integrate

[tex]W =15 \frac{x^2}{2}|\limits^{60}_{30}[/tex]

This gives:

[tex]W =15 *\frac{60^2 - 30^2}{2}[/tex]

[tex]W =15 *\frac{2700}{2}[/tex]

[tex]W =15 *1350[/tex]

[tex]W =20250N-cm[/tex]

Convert to Nm

[tex]W =\frac{20250Nm}{100}[/tex]

[tex]W =202.50Nm[/tex]

[tex]\sf\purple{A.\:Between \:12\:and\:13.}[/tex] ✅

[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]

[tex] \sqrt{156} \\ = 12.4899 \\ = 12.49[/tex]

Therefore, [tex] \sqrt{156} [/tex] will fall in between 12 and 13.

[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]

Answer:

The answer is 5 units.

Step-by-step explanation:

I just counted the squares

Answer:

133 fishes

Step-by-step explanation:

Units of food A = 400 units

Units of food B = 400 units

Fish Bass required 2 units of A and 4 units of B.

Fish Trout requires 5 units of A and 2 units of B.

i. For food A,

total units of food A required = 2 + 5

                                                 = 7 units

number of bass and trout that would consume food A = 2 x [tex]\frac{400}{7}[/tex]

                                               = 114.3

number of bass and trout that would consume food A = 114

ii. For food B,

total units of food B required = 4 + 2

                                                = 6 units

number of bass and trout that would consume food B = 2 x [tex]\frac{400}{6}[/tex]

                                                 = 133.3

number of bass and trout that would consume food B = 133

Thus, the maximum number of fish that the lake can support is 133.

Answer:

-25

Step-by-step explanation:

-24×(-25)=600

Hope this helps! :)

Answer: It's -25

edg 2023

Answer:

The correct option is (b).

Step-by-step explanation:

The solution of the given polynomial is :

[tex](-\dfrac{1}{3},4)[/tex]

x = 1/3 and y = -4

i.e.

Sum of roots = (1/3-4) = -11/3

Product of roots = (1/3)(-4) = -4/3

The quadratic equation is as follows :

[tex]x^2+(\text{sum of roots})x+\text{Product of roots}=0[/tex]

Put all the values,

[tex]x^2+\dfrac{-11}{3}x+\dfrac{-4}{3}=0\\\\3x^2-11x-4=0[/tex]

So, the correct option is (b).

Answer:

Step-by-step explanation:

From the information given:

Tank's Capacity = 600 gal

Original content of water = 200 gal

Salt solution = 100 lb

In the tank:

Suppose x(t) = amount of salt at time (t); &

V(t) = volume of water

Then,

V(t) = 200 + t

and [tex]x'(t) = 3 - \dfrac{2x(t)}{200+t}[/tex]

From the above linear equation, the integrating factor can be computed as:

[tex]x(t) = \Bigg(e^{\int \dfrac{2}{200+t}} \Bigg)^{dt}[/tex]

[tex]= e^{2 \ log (200+t)}[/tex]

= (200 + t)²

The general solution can now be expressed as:

[tex]x(t) = \dfrac{1}{(200+t)^2}\Bigg( \int 3(200+t)^2 \ dt +C\Bigg)[/tex]

We know that C = integrating factor, thus taking the integral:

[tex]x(t) = \dfrac{(200+t)^3 +C}{(200+t)^2}[/tex]

At the initial condition, x(0) = 100

∴

[tex]100= \dfrac{(200)^3 +C}{(200)^2}[/tex]

C = ((200)²×100) - 200³

C = -4 × 10⁶

Hence, at any time t, the amount of salt is:

[tex]x(t) = \dfrac{(200+t^2) - 4\times 10^6}{(200+t)^2}[/tex]

Answer:

The answer is below

Step-by-step explanation:

The profit equation is given by:

p(t)= -25t³+625t²-2500t

The maximum profit is the maximum profit that can be gotten from selling t trailers. The maximum profit is at point p'(t) = 0. Hence:

p'(t) = -75t² + 1250t - 2500

-75t² + 1250t - 2500 = 0

t = 2.3 and t = 14.3

Therefore t = 3 trailers and t = 15 trailers

p(15) = -25(15³) + 625(15²) - 2500(15) = 18750

Therefore the company makes a maximum profit of approximately $18750 when it sells approximately 15 trailers.

Answer:

See below

Step-by-step explanation:

Since t is number of trailers, the domain includes only those values greater than 0.

On the relevant domain, the graph crosses the x-axis at the points (5,0) and (20,0). Between these points, the value of p(t) is positive. So the company makes a profit when it sells between 5 and 20 trailers.

On the positive interval between these points, the graph reaches a relative maximum when t roughly equals 14 and p(t) roughly equals $19,000.

So the maximum profit of approximately $19,000 occurs when it sells approximately 14 trailers.

Answer:

You should expect 4 of the next 20 children to enroll to be toddlers.

Step-by-step explanation:

This question is solved by proportions.

So far:

We have that of 2 + 8 = 10 children, 2 are toddlers, so the proportion of toddlers is 2/10 = 0.2.

How many of the next 20 children to enroll should you expect to be toddlers?

0.2 out of 20, so: 0.2*20 = 4

You should expect 4 of the next 20 children to enroll to be toddlers.

Answer:

k>1.25

Step-by-step explanation:

The given quadratic equation is :

(k – 1)x² + x + 1 = 0

We need to find all values of k for which the above equation has two real and unequal roots.

For a quadratic equation ax²+bx+c=0, for real and unequal roots,

b²-4ac>0

Here, a = (k-1), b = 1 and c = 1

Put all the values,

1²-4×(k-1)1>0

1-4k+4>0

5-4k>0

k>1.25

S, k can take values more than 1.25. Hence, it can take values 1.5, 1.75.

Answer:

0.64(x) = 30

Step-by-step explanation:

Hope that's correct.

Answer:

-10, -1

Step-by-step explanation:

See Image below:)