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ANTONII [103]
3 years ago
9

HELP ME PLEASE LOVE U <33

Mathematics
1 answer:
vazorg [7]3 years ago
3 0

i think the answer is 56.55

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68 - 4 × (6 - 4)4<br><br> pls help o-o
Alika [10]

Answer:

Here you go!, hope it helped. 68-32x

7 0
2 years ago
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Consider the following function. f(x) = 9 − x2/3 Find f(−27) and f(27). f(−27) = f(27) = Find all values c in (−27, 27) such tha
oksano4ka [1.4K]

I guess the function is f(x)=9-x^{2/3}. Then f(-27)=0 and f(27)=0.

The derivative is f'(x)=-\dfrac23 x^{-1/3}, but there is no c such that

-\dfrac23c^{-1/3}=0

This doesn't contradict Rolle's theorem because f'(0) does not exists. In other words, f is not differentiable on (-27, 27), so the conditions of Rolle's theorem are not met. (Looks like that would be the last option, or the second to last option if the last one is "Nothing can be concluded")

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3 years ago
Use the number line to add or subbtract integers please help me with this
Digiron [165]

Answer:

a)=+2 b)=1

Step-by-step explanation:

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8 0
3 years ago
A particle moves along the curve y = 5x^2 – 1 in such a way that the y value is decreasing at the rate of 2 units per second. At
Shtirlitz [24]

Answer:

The correct option is;

Increasing one fifth unit/sec

Step-by-step explanation:

The equation that gives the curve of the particle of the particle is y = 5·x² - 1

The rate of decrease of the y value dy/dt = 2 units per second

We have;

dy/dx = dy/dt × dt/dx

dy/dx = 10·x

dy/dt = 2 units/sec

dt/dx = (dy/dx)/(dy/dt)

dx/dt = dy/dt/(dy/dx) = 2 unit/sec/(10·x)

When x = 1

dx/dt = 2/(10·x) = 2 unit/sec/(10 × 1) = 1/5 unit/sec

dx/dt = 1/5 unit/sec

Therefore, x is increasing one fifth unit/sec.

8 0
3 years ago
Carlos is using the distributive property to evaluate the expression 14 (49) by using friendlier numbers. His work is shown belo
Natasha2012 [34]
Separating 49 into 40+9
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