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morpeh [17]
3 years ago
14

PLEASE HELPP 6TH GRADE MATH!!

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
8 0

Answer: 1- 2:3

2- 4:7

3- draw 5 of something, then one of something else

4- same as 3

5- 10, 3, and 20

6- 14, 21, and 4

Step-by-step explanation:

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Solve this indices a²b + a³b / a²b²​
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Step-by-step explanation:

Notice: The literal factors are all the combinations of a and b where the sum of the exponents is 4: a4, a³b, a²b², ab³, b4 ... The solution to the problem of the binomial coefficients without actually ... The upper index n is the exponent of the expansion; the lower index k indicates which term ... 1a5 + a4b + a3b² + a²b3 + ab4 + b5.

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3 years ago
Can anyone help me please? I have no idea what to do! :'(
liberstina [14]

Answer:

7) 10^(3/2)

8) 2^(1/6)

9) 2^(5/4)

10) 5^(5/4)

Step-by-step explanation:

7)  (√10)^3 = 10^(3/2)

8)  6 root 2 = 2^(1/6)

9)  (4 root 2)^5 = 2^(5/4)

10) (4 root 5)^5 = 5^(5/4)

5 0
3 years ago
What is 2+2 equals to ??
romanna [79]

Answer: 4

Step-by-step explanation:

5 0
2 years ago
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Three runners run along a 1.5 mile trail one Saturday
Reil [10]

Answer:need graph

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3 years ago
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A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
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