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den301095 [7]
3 years ago
7

PLEASE HELP QUICKLY!

Mathematics
1 answer:
tino4ka555 [31]3 years ago
6 0

Answer:

the answer should be d

Step-by-step explanation:

1 x 4 =4

2 x 4 =8

3 x 4 =12

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Identify the absolute extrema of the function and the x-values where they occur.
sladkih [1.3K]
f(x)=6x+\dfrac{24}{x^2}+3\\
f'(x)=6-\dfrac{48}{x^3}\\
6-\dfrac{48}{x^3}=0\\
6x^3-48=0\\
6x^3=48\\
x^3=8\\
x=2\\


For x the derivative is negative.
For x>2 the derivative is positive.
Therefore at x=2 there's a minimum.

f_{min}=6\cdot2+\dfrac{24}{2^2}+3=12+6+3=21

6 0
3 years ago
Please help me answer this with the correct answer :)
Yuki888 [10]

Answer: b

Step-by-step explanation:

5 0
3 years ago
30 POINTS!!!
Daniel [21]

The unit rate that corresponds to the proportional relationship shown in the given graph above is: 4/3 cm/s.

<h3>How to Find the Unit Rate of a Proportional Graph?</h3>

The unit rate of a proportional graph is determined using the formula below:

k = y/x, where x and y are coordinates of any point on the line.

Thus, to find the unit rate of a proportional graph, pick the coordinates of any point on the line and find k = y/x.

From the given graph, let's pick the indicated point on the line having the coordinates, (12,16). Find k:

Unit rate (k) = 16/12

Simplify

Unit rate (k) = 4/3

Therefore, the unit rate that corresponds to the proportional relationship shown in the given graph above is: 4/3 cm/s.

Learn more about proportional graph on:

brainly.com/question/23318486

#SPJ1

8 0
1 year ago
HELPPPP BRAINLIEST I WILL GIVE
Leto [7]

Answer:

b

Step-by-step explanation:

i think

7 0
3 years ago
GIVING BRAINLIESTTTTT
andreev551 [17]

Answer:

D.Place the compass on a point on the original line.

Complete Question:

A.Create only one arc that intersects the original line.

B.Create two arcs that intersect the original line.

C.Place the compass on a point off the original line.

D.Place the compass on a point on the original line.

Step-by-step explanation:

* Lets revise the steps of constructing parallel lines and a perpendicular line through a point on the line

# Contracting parallel lines

- Given: Line AB and point P not on AB

1. Draw a slant line through point P and intersects AB at point C, this line is the transversal of the parallel lines

2. Place the pin of the compass at point C and draw an arc intersects AB at D and CP at E

3. Without changing the distance of the compass put the pin of the compass on point P and draw an arc intersects CP at point Q

4. Use the compass to measure the distance from D to E and place the pin of the compass at point Q and draw an arc intersects the arc from P to CP at point U

5. Join P and U by a line this line is parallel to AB

# Constructing a perpendicular line through a point on the line

- Given line AB and point P lies on it

1. Place your compass pin at P and draw an arc of any size below AB that crosses the line twice at points C and D

2. Stretch the compass to a larger distance

3. Place the compass pin on point C and draw a small arc above the line

4. Without changing the distance of the compass place the compass pin on point D and draw another arc intersects the arc from C at point E

5. Using a straightedge, join P and E where PE is perpendicular to AB

* Lets find the same steps in the constructions

∵ In first construction we put the pin of the compass at point C which  lies in the original line AB

∵ In second construction we put the pin of the compass at point P which lies in the original line AB

∴ The common step is:  "Place the compass on a point on the original line"

6 0
3 years ago
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