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Phantasy [73]
3 years ago
15

Students in a science class are divided into 6 equal groups with at least 4 students in each group for a project. Write and solv

e an inequality that represents the number of students in the class
Mathematics
2 answers:
natta225 [31]3 years ago
6 0
Is that right ^ i need help to
Tatiana [17]3 years ago
3 0
Since the glass is divided into 6 groups and 4 in each you multiply both of them to find the least number of students in the class.
6(4)<x
24<x
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Please help me solve this!!
Brut [27]

Answer:

30

by-step explanation:

5*5=30

3 0
3 years ago
1. Writing an equation for an exponential function by
Naya [18.7K]

Answer: 1) t(n)=0.6(2)^n

2) f(x)=10(5)^x

Step-by-step explanation:

1) Let the function that shows the thickness of the paper after n folds,

t(n) = ab^n         ---------(1)

Since, According to the question,

Initially the thickness of the paper = 0.6

That is, at n = 0, t(0) = 0.6

By equation (1),

0.6 = a(b)^0\implies 0.6 = a

Hence the function that shows the given situation,

t(n) = 0.6 b^n       -----------(2)

Again when we fold the paper the thickness of the paper will be doubled.

Thus, at n = 1, t(1) = 1.2

By equation (2),

1.2 = 0.6 b^1\implies 2 = b

Thus, the complete function is,

t(n) = 0.6 (2)^n    

2) Let the function that is passing through the points (-2, 2/5) and (-1,2),

f(x) = ab^x         ---------(1)

For f(x) = 2, x = -1

By equation (1),

2= ab^{-1}       ---------(2)

Also, For f(x) = 2/5, x = -2

Again, By equation (1),

\frac{2}{5}= a(b)^{-2}

\implies \frac{2}{5}=ab^{-1}b^{-1}=2b^{-1}

\implies \frac{2}{5}=\frac{2}{b}

\implies 2b=10

\implies b = 5

By substituting this value in equation (2),

We get, a = 10

Hence, from equation (1), the function that is passing through the points (-2, 2/5) and (-1,2),

f(x) = 10(5)^x

5 0
3 years ago
Factor a perfect square
Lesechka [4]
(5x-12)^2

Square factor of 25 is 5.
Square factor of 144 is 12.

Hope this helps. I didn't understand what to do with it.
5 0
3 years ago
Write the ration for sin A and cos A
Archy [21]

sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}

\sin A=\dfrac{3}{5},\ \cos A=\dfrac{4}{5}

6 0
3 years ago
(x-3) to the 2 = 0 in standard form​
Ivenika [448]
X-3+2 is the answer to this question
3 0
3 years ago
Read 2 more answers
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