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2 years ago

Craig drank two 5-cup bottles of water in one day how many fluid onces is that

2 answers:

5 0

One cup = (is) 8 ounces. So if he had 5 cups you must multiply 8 x 5 and get 40. So Craig drank 40 ounces in one day. :)

Sever21 [200]2 years ago

3 0

I got 80 fluid ounces
2 × 5=10 which would be how many cups he drank
and one cup is 8 fluid ounces
so 10×8= 80 ounces which is all he drank that day

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F(x)=2x and g(x)=square root of x what is the domain of f(g(x))?

Dmitriy789 [7]

Answer:

$x\ge 0$

Step-by-step explanation:

The given functions are:

$f(x)=2x$

and

$g(x)=\sqrt{x}$

Now,

$f(g(x))=f(\sqrt{x} )$

$f(g(x))=2\sqrt{x}$

The domain refers to values of $x$ for which $f(g(x))=2\sqrt{x}$ is defined.

The expression under the radical sign must not be negative.

The domain is

$x\ge 0$

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2 years ago

Solve 2 | 2x + 10|+ 4 > 10

Valentin [98]

Answer:

x>−72 or x<−132

Step-by-step explanation:

I believe that is the answer

8 0

3 years ago

Solve the following proportion for x. 3/13 = x/5 Round your answer to the nearest tenth.

anastassius [24]

To solve, isolate the x. Cross multiply

3/13(13)(5) = x/5(5)(13)

3(5) = (13)x

Simplify

13x = 15

Isolate the x. Divide by 13.

13x/13 = 15/13

x = 15/13

x = 1.2

1.2 is your answer

hope this helps

8 0

2 years ago

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

2 years ago

What is the inverse of the function f(x) = 2x – 10?

Nadusha1986 [10]

Answer:

f⁻¹(x) = (1/2)x +5

Step-by-step explanation:

In y = f(x), swap the variables, then solve for y. The expression you get is f⁻¹(x).

... y = 2x -10

... x = 2y -10 . . . . . . swapped variables

... x +10 = 2y . . . . . add 10

... (1/2)x + 5 = y . . . . divide by 2

... f⁻¹(x) = (1/2)x + 5 . . . . . . rewrite using function notation

6 0

3 years ago