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Oksana_A [137]
3 years ago
10

Please help I will mark you brainliest​

Mathematics
1 answer:
ddd [48]3 years ago
8 0
Kbivivivogxoyxoycoycoycoycoycoycoycoyco
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A recipe for a loaf of bread requires 2/3 cup of flour. How many cups of flour would it take to make 15 loaves of bread?
Mars2501 [29]

Answer:

10

Step-by-step explanation:

1 loaf if bread= 2/3 cup

15 loaves of bread= 2/3 × 15

= 10

5 0
2 years ago
A fair die is rolled in a board game to determine how many spaces a player can move on their turn. It has one of the numbers 1,2
Jet001 [13]

Answer:

(a)6 (b)3 (c)0.5

Step-by-step explanation:

Given a fair die with the numbers 1,2,3,4,5, or 6 on each of its faces.

  • Event A is the event of rolling an even number
  • Event B is the event of rolling an odd number.

(a)The sample space for the outcomes in this experiment is {1,2,3,4,5,6}

There are <u>6</u><u> </u>outcomes in the sample space.

n(S)=6

(b)

Event A is the event of rolling an even number

Sample space of A = {2,4,6}

There are <u>3</u> outcomes in event A.

n(A)=3

(c)The probability of event A

P(A)=\dfrac{n(A)}{n(S)} =\dfrac{3}{6} =0.5

P(A) =<u>0.5</u> is the probability that you choose an even number.

6 0
2 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
Read 2 more answers
The expression describe the number that is 8 to the left of -8 on the number line ​
mart [117]

Start at -8 move to the left 8 more times,. You get -16.

4 0
3 years ago
Can anyone help with only #1 please, will give you brainliest!!
Tresset [83]

No, it is <u>not</u> a translation. The arrow starts pointing to the right and then points upward. A translation would have the arrow always pointing right the entire time, only that its position would move.

====================================================

Problem 2

We can shift the K up 6 units to get what you see in the attached image below

5 0
3 years ago
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