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Maksim231197 [3]
3 years ago
12

Problem 1.3

Mathematics
1 answer:
Bess [88]3 years ago
6 0

Answer:

Get a brain

Step-by-step explanation:

Be smart

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members of a soccer team raised $741.15 to go to a tournament. They rented a bus for $250 and budgeted $25.85 per player for mea
blsea [12.9K]

9514 1404 393

Answer:

  19 players

Step-by-step explanation:

The total being spent is ...

  bus cost + (per player cost) × (number of players) = total raised

  250 + 25.85p = 741.15

  25.85p = 491.15

  p = 491.15/25.85 = 19

The team could bring 19 players to the tournament.

6 0
2 years ago
If there are 5 terms in a quadratic sequence
exis [7]

Answer:

7.2

Step-by-step explanation:

First differences:

0.8-0.2, 1.8-0.8, 3.2-1.8, 5.0-3.2

0.6, 1.0, 1.4, 1.8

Second differences:

1.0-0.6, 1.4-1.0, 1.8-1.4

0.4, 0.4, 0.4

For the next term, add 0.4 to the last term from the first differences, and add the sum to the last term of the quadratic sequence:

0.4+1.8=2.2

2.2+5.0=7.2

∴ The next term in the sequence is 7.2

7 0
3 years ago
Julie wanted to earn money to attend a basketball camp. She started with a gift of money from her grandparents and then saved $2
adoni [48]
The equation would be y = 25x + 50

In order to find this equation, we need to know how much the gift from her grandparents was. To do so, we have to find out how much she's saved from dog walking. 

Since she saves $25 a month for 7 months, we can find the total amount as:

25*7 = 175

Then we can subtract that from the total she has saved to find the amount for the gift.

225 - 175 = 50

Finally, we put the amount per month in the equation with the gift as the y intercept to create the equation above. 
6 0
3 years ago
F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte
Mekhanik [1.2K]

Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

(b) Local minimum at (1, 0)

(c) No inflection point; concave up on (0, ∞)

Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

           2x² - x  -1 = 0

     ( x - 1) (2x + 1) = 0

         x = 1 or x = -½

We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½  = ⁵/₂

ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

             x² + 2 = 0

                   x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

\begin{array}{lccc}\text{Test} & x < 1 & x = 1 & x > 1\\\text{Sign of f''} & + & 0 & +\\\text{Concavity} & \text{up} & &\text{up}\\\end{array}

The function is concave up on (0, ∞).

6 0
3 years ago
A company finds that on average, they use 43 minutes for each meeting.
lilavasa [31]
F. 220
Hope this helps!
3 0
3 years ago
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