HELP ASAP TIMED!!!!!!!!!!!!!!!

PLEASE ANSWER ASAP WILL MARK BRAINLIEST!!!

lys-0071 [83]

Answer:

C. Never

Step-by-step explanation:

The legs of a right triangle are never longer than the hypotenuse. Otherwise, the legs would never connect to the hypotenuse.

3 0

3 years ago

Read 2 more answers

A product can be made in sizes huge, average and tiny which yield a net unit profit of $14, $10, and$5, respectively. Three cent

navik [9.2K]

Answer:

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃ to maximize

Subject to:

First center X₁₁ + X₂₁ + X₃₁ ≤ 550

Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

X₁₁ + X₁₂ + X₁₃ ≤ 710

X₂₁ + X₂₂ + X₂₃ ≤ 900

X₃₁ + X₃₂ + X₃₃ ≤ 350

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

Step-by-step explanation:

Let´s call Xij product size i produced in center j

According to this, we get the following set of variable

X₁₁ product size huge produced in center 1

X₁₂ product size huge produced in center 2

X₁₃ product size huge produced in center 3

X₂₁ product size average produced in center 1

X₂₂ product size average produced in center 2

X₂₃ product size average produced in center 3

X₃₁ product size-tiny produced in center 1

X₃₂ product size-tiny produced in center 2

X₃₃ product size-tiny produced in center 3

Then Objective function is

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Constrains

Center capacity

1.- First center X₁₁ + X₂₁ + X₃₁ ≤ 550

2.- Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

3.- Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

Water available

1.- 22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

2.- 22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

3.- 22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

Demand constrain

Product huge

X₁₁ + X₁₂ + X₁₃ ≤ 710

Product average

X₂₁ + X₂₂ + X₂₃ ≤ 900

Product tiny

X₃₁ + X₃₂ + X₃₃ ≤ 350

Fraction SP/CC must be the same

First and second centers fraction SP/CC

(X₁₁ + X ₂₁ + X₃₁)/ 11000 = (X₁₂ + X₂₂ + X₃₂)/ 2700

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

First and third centers fraction SP/CC

(X₁₁ + X ₂₁ + X₃₁)/ 11000 = ( X₁₃ + X₂₃ + X₃₃)/ 3400

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Subject to:

First center X₁₁ + X₂₁ + X₃₁ ≤ 550

Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

X₁₁ + X₁₂ + X₁₃ ≤ 710

X₂₁ + X₂₂ + X₂₃ ≤ 900

X₃₁ + X₃₂ + X₃₃ ≤ 350

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

6 0

3 years ago

Solve each equation by completing the square <br>r² = –12r - 35

Fittoniya [83]

Answer:

r = - 7, r = - 5

Step-by-step explanation:

Given

r² = - 12r - 35 ( add 12r to both sides )

r² + 12r = - 35

To complete the square

add ( half the coefficient of the r- term )² to both sides

r² + 2(6)r + 36 = - 35 + 36

(r + 6)² = 1 ← take the square root of both sides )

r + 6 = ± 1 ( subtract 6 from both sides )

r = - 6 ± 1, thus

r = - 6 - 1 = - 7 or r = - 6 + 1 = - 5

7 0

3 years ago

Solve for x: 21x - 3| + 1 = 7

Andreas93 [3]

Answer:

x=6

Step-by-step explanation:

distribute 2 to x and 3..you get 2x -6 +1. add 6 and subtract one to cancel it out and bring it to the other side. 2x=12...simplify x=6

4 0

3 years ago

Question is in the picture, please help me out!!

Grace [21]

The arithmetic sequences are as follow:

<h3>What is Arithmetic Sequence?</h3>

An arithmetic sequence in algebra is a sequence of numbers where the difference between every two consecutive terms is the same.

1) t(n) = 5n + 4

t(1) = 9, t(2) = 14, t(3) = 19

So,

9,14,19,...

d= 14-9 = 5

d= 19-14 =5

Hence, it is an AP

2) 1, 2, 4, 8 , 16

Hence, it is not an AP

3) 3, 6, 9 ,...

It is an AP

4)It is given that it is an AP

5) tn = 2*3^n

t1= 6, t2= 18, t3= 54

So, 6, 18, 54,...

Hence, it is not an AP

6) 3 , 1, 1/3,...

It is not an AP

7) t(n+1)= 6*t(n)

t(1) = -1

t(2)=-6

t(3)= -36

Hence, it is not an AP

8) -3, 1, 5, 9

Hence, it is an AP.

9) 1, 4, 9,...

Hence, it not an AP

10) 2,1,0,1,2,...

It is not an AP

11) t(n)= -2n-5

t(1)= -7, t(2)= -9, t(3)= -11

Hence, it is an AP

12) tn= (1/2)^n

t1= 1/2, t2= 1/4, t3= 1/8

It is not an AP

Learn more about AP here:

brainly.com/question/24873057

#SPJ1

6 0

2 years ago