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g100num [7]
3 years ago
9

Paul spends $89.10 on blackberries. If the blackberries cost $4.95 per pound, how many pounds of blackberries did he buy?

Mathematics
1 answer:
aksik [14]3 years ago
4 0
18 pounds
U can divide 89.10 by 4.95 and u get 18
To get if it’s right multiple 18 by 4.95 and U get 89.10
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aniked [119]

Answer:

A. Plane B because it was 9.33 miles away

B. 48 units

Step-by-step explanation:

A. Since the airplanes fly at an angle to the runway, their direction forms a triangle with the runway with their height above the ground as the opposite of the angle and their distance from the airport as the hypotenuse.

So for airplane A with 44° angle of departure,

sin44° = y/h where y = height above the ground and h = distance from airport

So h = y/sin44° = 6/sin44° = 8.64 miles

So for airplane B with 40° angle of departure,

sin40° = y/H where y = height above the ground and H = distance from airport

So H = y/sin40° = 6/sin40° = 9.33 miles

Since airplane B is at 9.33 miles away from the airport whereas airplane A is 8.64 miles from the airport, airplane B is farther away.

B. We know that scale factor = new size/original size

Our scale factor = 4 and original size = 12 units. So,

new size = scale factor original size = 4 × 12 = 48 units.

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3 years ago
The expression is 12n -4 = -7 solve for n
VARVARA [1.3K]
The answer for N is 1/4
6 0
3 years ago
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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, ∞).

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tangare [24]
Answer
8

Explanation

The absolute value of a negative number turns into a positive
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C. 17

Why:

Divide 100 by 6, round the answer and you should get 17

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