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coldgirl [10]
2 years ago
11

Which algebraic expression represents the word expression "three times the square of a number, n, subtracted from 54"?

Mathematics
1 answer:
sveta [45]2 years ago
3 0

Answer:

54 - 3n²

Step-by-step explanation:

Square of a number n = n²

Three times the square of n = 3*n² = 3n²

Expression:

54 - 3n²

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The length of a rectangle is three times its width the length is 3 3/4 feet. What is the area if the rectangle?
Roman55 [17]

Answer:

A = 4 11/16 or 4.6875

Step-by-step explanation:

3 3/4 = 3W

divide both sides by 3

15/4 divided by 3 = W

15/4 x 1/3 = W

5/4 = W

A = L x W

3 3/4 x 1 1/4 = 15/4 x 5/4 = 75/16 = 4 11/16

4 0
2 years ago
If one check was written and one deposit was made into an account in one day, the account’s ending balance on that day _________
AlekseyPX
Hello! I believe the correct answer is C.


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3 years ago
What number am i missing please only tell me the one missing!!!
makvit [3.9K]

Answer:

16

Step-by-step explanation:

40 can be multiplied by 5, 8 times. 8 multiplied by 2 is 16.

4 0
2 years ago
(-2x+y=8) minus (5x-y=6)
Hitman42 [59]

Answer:-7x+2y=2

Step-by-step explanation:

(-2x+y=8)-(5x-y=6)

(-2x+y=8)> this side change since it’s negative

(-2x+y=8) - (-5x+y=-6)

-7x+2y=2

6 0
2 years ago
Read 2 more answers
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
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