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Sergeeva-Olga [200]
3 years ago
9

Find the location of the absolute maximum and absolute minimum of the function on the interval [ -4,0).

Mathematics
1 answer:
N76 [4]3 years ago
3 0
Maximum is 0 and minimum is -4. Due to the brackets, that means that this is where the value terminates. The parentheses next to the zero means equal to
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What is the simplified form?
yarga [219]

Answer:

\frac{6}{5x^{10}}

Step-by-step explanation:

Step 1: Simplify square roots

√72 = 6√2

√50 = 5√2

So, \frac{x^86\sqrt{2} }{x^{18}5\sqrt{2} }

Step 2: Cancel like terms

\frac{6}{x^{10}(5)}

Step 3: Rewrite

And we should get our answer!

7 0
3 years ago
Prove that the altitude to the base of an isosceles triangle is also the median to the base
LenKa [72]

The base angles theorem converse states if two angles in a triangle are congruent, then the sides opposite those angles are also congruent. The Isosceles Triangle Theorem states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle.

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4 0
3 years ago
-10 + 20p + 6p = -16
pshichka [43]

Answer: p= -3/13

Step-by-step explanation:

-10+26p=-16

+10 +10

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26 26

p= -3/13

6 0
3 years ago
Consider an object moving along a line with the given velocity v. Assume t is time measured in seconds and velocities have units
Yakvenalex [24]

Answer:

a) positive direction: t < 2s & t>6s   ; negative direction: 2s < t < 6s

b) 7 m

c) 71 m

Step-by-step explanation:

Given:

v(t) = 3t^2 -24t +36   [0 , 7]

Find:

a) The value of time when particle is moving in positive direction:

The change in direction of the particle can be determined by v(t) > 0

Hence,

                                    0 < 3t^2 -24t +36

                                    0 < t^2 - 8t + 12

                                    0 < (t - 2)*(t - 6)

                                    t < 2s  , t  > 6s  

The particle travels in positive direction in the interval t < 2s and t > 6s , While it travels in negative direction when 2s < t < 6s.

b) The displacement ds over the given interval [ 0 , 7 ]

                                  ds = integral (v(t)).dt

                                  ds = t^3 -12t^2 +36t

                                  ds = 7^3 -12*7^2 +36*7

                                  ds = 7 m

c) Total distance traveled in the interval:

                                 Total distance= ds(0-2) + ds(2-6) + ds(6-7)

                                 D = 2*(2^3 -12*2^2 +36*2) - 2*(6^3 -12*6^2 +36*6) + 7

                                 D = 2*32 - 2*0 + 7

                                 D = 71 m          

     

6 0
3 years ago
PLZZZZZ HELPPP ASAPPPP IM FREAKIN OUTTTTTT
Temka [501]

Answer:

1. C)  2 (x-2)(x-1)

2. D) cannot be factored

3. B)  (x-8)(x+5)

4. D)  2 (x+5)(x-5)

5. D)  (x+5)(x-3)

hope this was helpful! let me know if you needd the steps! c:

3 0
3 years ago
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