-4+2(12) ———— -3(-3)-(-1)

Prove the following using a direct proof:

hodyreva [135]

Answer:

Step-by-step explanation:

Prove: That the sum of the squares of 4 consecutive integers is an even integer.

An integer is a any directed number that has no decimal part or indivisible fractional part. Examples are: 4, 100, 0, -20,-100 etc.

Selecting 4 consecutive positive integers: 5, 6, 7, 8. Then;

$5^{2}$ = 25

$6^{2}$ = 36

$7^{2}$ = 49

$8^{2}$ = 64

The sum of the squares = 25 + 36 + 49 + 64

= 174

Also,

Selecting 4 consecutive negative integers: -10, -11, -12, -13. Then;

$-10^{2}$ = 100

$-11^{2}$ = 121

$-12^{2}$ = 144

$-13^{2}$ = 169

The sum of the squares = 100 + 121 + 144 + 169

= 534

Therefore, the sum of the squares of 4 consecutive integers is an even integer.

8 0

3 years ago

Help xd -(x+2) = 3(7x - 8)= 7(x+3)-5x Thank you!

HACTEHA [7]

Answer:

x=−5

x= 31/21 =1 10/21= 1.476190476

2x+21

hope this helps

3 0

3 years ago

Read 2 more answers

Whats the answer to this???

garri49 [273]

Answer:

A: 2/5

B: -0.2

C: -4/5

4 0

2 years ago

Apple launches an ad campaign for iPhone 12. Managers want to know what is the % of social media mentions that have positive rea

Orlov [11]

Answer:

social media mentions do managers at Apple need to collect, if they want to get a margin of error of 2%

Sample size 'n' = 625

Step-by-step explanation:

Step(i):-

Given data Apple launches an ad campaign for iPhone 12

Managers want to know what is the % of social media mentions that have positive reactions to the new ad campaign

probability of positive reactions p = 1/2

Probability of negative reactions q = 1-p = 1/2

we know that $\sqrt{p(1-p}\leq \frac{1}{2}$

Step(ii):-

Given margin of error = 2 % =0.02

The margin of error is determined by

$M.E = \frac{\sqrt{p(1-p} }{\sqrt{n} }$

$0.02 = \frac{\frac{1}{2} }{\sqrt{n} }$

$\sqrt{n} =\frac{1}{2 X0.02}$

√n = 25

squaring on both sides, we get

n = 625

8 0

3 years ago

A cylindrical tank has a base of diameter 12 ft and height 5 ft. The tank is full of water (of density 62.4 lb/ft3).(a) Write do

saw5 [17]

Answer:

a. 71884.8 π lb/ft-s²∫₀⁵(9 - y)dy

b. 23961.6 π lb/ft-s²∫₀⁵(5 - y)dy

c. 99840π lb/ft-s²∫₀⁶rdr

Step-by-step explanation:

.(a) Write down an integral for the work needed to pump all of the water to a point 4 feet above the tank.

The work done, W = ∫mgdy where m = mass of cylindrical tank = ρA([5 + 4] - y) where ρ = density of water = 62.4 lb/ft³, A = area of base of tank = πd²/4 where d = diameter of tank = 12 ft.( we add height of the tank + the height of point above the tank and subtract it from the vertical point above the base of the tank, y to get 5 + 4 - y) and g = acceleration due to gravity = 32 ft/s²

So,

W = ∫mgdy

W = ∫ρA([5 + 4] - y)gdy

W = ∫ρA(9 - y)gdy

W = ρgA∫(9 - y)dy

W = ρgπd²/4∫(9 - y)dy

we integrate W from y from 0 to 5 which is the height of the tank

W = ρgπd²/4∫₀⁵(9 - y)dy

substituting the values of the other variables into the equation, we have

W = 62.4 lb/ft³π(12 ft)² (32 ft/s²)/4∫₀⁵(9 - y)dy

W = 71884.8 π lb/ft-s²∫₀⁵(9 - y)dy

.(b) Write down an integral for the fluid force on the side of the tank

Since force, F = ∫PdA where P = pressure = ρgh where h = (5 - y) since we are moving from h = 0 to h = 5. So, P = ρg(5 - y)

The differential area on the side of the tank is given by

dA = 2πrdy

So. F = ∫PdA

F = ∫ρg(5 - y)2πrdy

Since we are integrating from y = 0 to y = 5, we have our integral as

F = ∫ρg2πr(5 - y)dy

F = ∫ρgπd(5 - y)dy since d = 2r

substituting the values of the other variables into the equation, we have

F = ∫₀⁵62.4 lb/ft³π(12 ft) × 32 ft/s²(5 - y)dy

F = 23961.6 π lb/ft-s²∫₀⁵(5 - y)dy

.(c) How would your answer to part (a) change if the tank was on its side

The work done, W = ∫mgdr where m = mass of cylindrical tank = ρAh where ρ = density of water = 62.4 lb/ft³, A = curved surface area of cylindrical tank = 2πrh where r = radius of tank, d = diameter of tank = 12 ft. and h = height of the tank = 5 ft and g = acceleration due to gravity = 32 ft/s²

So,

W = ∫mgdr

W = ∫ρAhgdr

W = ∫ρ(2πrh)hgdr

W = ∫2ρπrh²gdr

W = 2ρπh²g∫rdr

we integrate from r = 0 to r = d/2 where d = diameter of cylindrical tank = 12 ft/2 = 6 ft

So,

W = 2ρπh²g∫₀⁶rdr

substituting the values of the other variables into the equation, we have

W = 2 × 62.4 lb/ft³π(5 ft)² × 32 ft/s²∫₀⁶rdr

W = 99840π lb/ft-s²∫₀⁶rdr

7 0

3 years ago