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goldfiish [28.3K]
2 years ago
8

How much will you get paid if you work 28.5 hours for $7.75 per hour?

Mathematics
2 answers:
andrey2020 [161]2 years ago
8 0

Answer:

$220.875

Step-by-step explanation:

For 1 hour you get $7.75

To find how much money you make in 28.5 hours... you multiply the number of hours with the number of dollars.

28.5 * 7.75

$220.875

Alexxandr [17]2 years ago
4 0
The Answer is $220.88
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Can someone help i need step by step so i can understand. thank you​
kati45 [8]

Answer:

5

Step-by-step explanation:

The answer is 5 because as you can see: 10/8, if you divide the 10 by 2 that would be 5. Same with the denominator, 8 divided by 2 is 4. So, your answer is 5/4. Or 5.

8 0
2 years ago
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Help please show the work if you can
Vlada [557]

Answer:

The inverse function of 3x - 4 is (x+4)/3. so a

Step-by-step explanation:

f(x) = 3x - 4

y = 3x - 4 replace f(x) with y

x = 3y - 4 replace x with y and y with x.

3y = x + 4 solve for y

y = (x+4)/3 replace y with f-1(x)

f-1(x) = (x+4)/3

8 0
3 years ago
What is the upper bound of the function f(x)=4x4−2x3+x−5?
inessss [21]

Answer:

(no global maxima found)

Step-by-step explanation:

Find and classify the global extrema of the following function:

f(x) = 4 x^4 - 2 x^3 + x - 5

Hint: | Global extrema of f(x) can occur only at the critical points or the endpoints of the domain.

Find the critical points of f(x):

Compute the critical points of 4 x^4 - 2 x^3 + x - 5

Hint: | To find critical points, find where f'(x) is zero or where f'(x) does not exist. First, find the derivative of 4 x^4 - 2 x^3 + x - 5.

To find all critical points, first compute f'(x):

d/( dx)(4 x^4 - 2 x^3 + x - 5) = 16 x^3 - 6 x^2 + 1:

f'(x) = 16 x^3 - 6 x^2 + 1

Hint: | Find where f'(x) is zero by solving 16 x^3 - 6 x^2 + 1 = 0.

Solving 16 x^3 - 6 x^2 + 1 = 0 yields x≈-0.303504:

x = -0.303504

Hint: | Find where f'(x) = 16 x^3 - 6 x^2 + 1 does not exist.

f'(x) exists everywhere:

16 x^3 - 6 x^2 + 1 exists everywhere

Hint: | Collect results.

The only critical point of 4 x^4 - 2 x^3 + x - 5 is at x = -0.303504:

x = -0.303504

Hint: | Determine the endpoints of the domain of f(x).

The domain of 4 x^4 - 2 x^3 + x - 5 is R:

The endpoints of R are x = -∞ and ∞

Hint: | Evaluate f(x) at the critical points and at the endpoints of the domain, taking limits if necessary.

Evaluate 4 x^4 - 2 x^3 + x - 5 at x = -∞, -0.303504 and ∞:

The open endpoints of the domain are marked in gray

x | f(x)

-∞ | ∞

-0.303504 | -5.21365

∞ | ∞

Hint: | Determine the largest and smallest values that f achieves at these points.

The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:

The open endpoints of the domain are marked in gray

x | f(x) | extrema type

-∞ | ∞ | global max

-0.303504 | -5.21365 | global min

∞ | ∞ | global max

Hint: | Finally, remove the endpoints of the domain where f(x) is not defined.

Remove the points x = -∞ and ∞ from the table

These cannot be global extrema, as the value of f(x) here is never achieved:

x | f(x) | extrema type

-0.303504 | -5.21365 | global min

Hint: | Summarize the results.

f(x) = 4 x^4 - 2 x^3 + x - 5 has one global minimum:

Answer: f(x) has a global minimum at x = -0.303504

5 0
3 years ago
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Can someone help me.​
cestrela7 [59]

Answer:

X=11

Step-by-step explanation:

To find x-intercept/zero substitute h(x)=0

0=x-11 move the variable to the left hand side and change it's sign

-x=-11 change the signs on both sides of the equation

x=11

4 0
3 years ago
Someone solve this please like NOW 4(5x-7)+12=16x+24
Bingel [31]

20x-28+12=16x+24 (i distributed the 4 to get this)

20x-16=16x+24 (added -28 & 12)

4x=40 (Subtracted 16x from both sides and added 16 to both sides)

(divide by 4 on both sides)

x=10

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