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postnew [5]
3 years ago
14

Find the GCF of each pair of monomials. 15x, 12x

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
6 0

Answer: your answer should be 3x

Step-by-step explanation:

Hope this helps :)

Leno4ka [110]3 years ago
4 0

Answer:

3x

Step-by-step explanation:

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The equation for line A is y = -2/3 x -4. Line A and Line B are perpendicular and the point ( -2 , 1) lies on Line B. Write the
gavmur [86]

A23425

Step-by-step explanation:

5 0
2 years ago
HeLP ME!!! 2/5w-2=7/5w solve the equation using the addition property of equality.
dangina [55]

The answer would be -2 i had the same problem on my quiz and the correct answer was -2

4 0
3 years ago
Really need help on this too ASAP thanks for whoever helps me
geniusboy [140]

Answer:

D 45

Step-by-step explanation:

90/2 = 45

135/3 = 45

180/4 = 45

4 0
2 years ago
A volleyball player sets the ball in the air, and the height of the ball after t seconds is given in feet by h= -16^2+12t+6. A t
JulsSmile [24]

h=-16²+12t+6

8=-256+12t+6

256-6+8=12t

-258=12t

t=-258/12

t=21.5

7 0
3 years ago
You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.
postnew [5]

<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

<u>Derivatives</u>

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

1)

The function provided in the question is

h(x)=2x^2-12x+20

Let's find the first derivative

h'(x)=4x-12

solving h'=0:

4x-12=0

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

2)

To find when will the ball land, we set h=0

2x^2-12x+20=0

Simplifying by 2

x^2-6x+10=0

Completing squares

x^2-6x+9+10-9=0

Factoring and rearranging

(x-3)^2=-1

There is no real value of x to solve the above equation, so the ball will never land.

This problem seems to be wrong because no minimum point was found and no point of landing exists

3 0
3 years ago
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