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Cloud [144]
3 years ago
15

Mean 60 points if all is correct, really need some help. As its not allowing 40/5= 8 mean.

Mathematics
1 answer:
Gala2k [10]3 years ago
5 0

Answer:

2^K≥40

2^5≥20

24/5 =4.75

40 books= 40/5 = 8 mean

9 2's

10 7's

13 12's

2 17's

6 22's

18+70+156+34+132= 410/40 =10.25  

150+156+70=306+70=376/40 books bought = 9.4 mean

Step-by-step explanation:

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Determine whether F(x)=4x^2-16x+6 has a maximum or a minimum value and find that value
Reil [10]

Answer:

f(x) has a minimum value of -10.

Step-by-step explanation:

It will have a minimum value because the coefficient of x^2 is positive.

To find its value we convert to vertex form:

f(x) = 4x^2 - 16x + 6

     = 4(x^2 - 4x) + 6

      = 4[(x - 2)^2 - 4] + 6

      = 4(x - 2)^2 - 16 + 6

      = 4(x - 2)^2 - 10.

So the minimum value is -10.

5 0
2 years ago
How do you do this plzz help
Lesechka [4]

the answer is 4.2 and i check if the answer is correct

and it was correct  

8 0
4 years ago
I don't this question so I need help with this problem
sveta [45]
So first divide 7,500 by 150 and you will get 50.Next,do 17X50 which will give you 850.The answer is 850 Gold beads.Hope that helped.Have a happy new year
4 0
3 years ago
5. The fourth graders are going to the
Llana [10]

Answer:

14 rows. Hope this helped :) !

5 0
3 years ago
Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t > 0.
Margarita [4]

Answer:

y = 3sin2t/2 - 3cos2t/4t + C/t

Step-by-step explanation:

The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt

Comparing the standard form with the given differential equation.

p(t) = 1/t and q(t) = 3cos(2t)

I = e^∫1/tdt

I = e^ln(t)

I = t

The general solution for first a first order DE is expressed as;

y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.

yt = ∫t(3cos2t)dt

yt = 3∫t(cos2t)dt ...... 1

Integrating ∫t(cos2t)dt using integration by part.

Let u = t, dv = cos2tdt

du/dt = 1; du = dt

v = ∫(cos2t)dt

v = sin2t/2

∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt

= tsin2t/2 - cos2t/4 ..... 2

Substituting equation 2 into 1

yt = 3(tsin2t/2 - cos2t/4) + C

Divide through by t

y = 3sin2t/2 - 3cos2t/4t + C/t

Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t

3 0
3 years ago
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