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dezoksy [38]
3 years ago
14

Help me in this

Mathematics
1 answer:
strojnjashka [21]3 years ago
5 0

Hello from MrBillDoesMath!

Answer:    C)


Discussion:

To solve x^2 = 36/81  take the square root of both sides.

x = square root of (36/81)

As 36 = 6*6 and 81 = 9*9,

x = square root of (6*6)/ (9.9) or 6/9  

But 6 and 9 share the common factor "3" so

x =  6/9 = (2*3)/(3*3) = 2/3              

The last equality was gotten by cancelling the common factor 3.

         

Regards, MrB

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A 2-column table with 7 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3. The sec
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The statements that are true about the intervals of the continuous function are Options 2, 4  and 5

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<h3>What is the statement about?</h3>

Looking at the values given, the intervals which satisfies the condition are known to be:

f(x)<=0 over the interval [0,2]

f(x)>0 over the interval (-2,0)

f(x)>=0 over the interval [2,∞)

Because:

Since the table with x  and f(x) values, we have to examine analyze the table and see the each option that is in line with f(x) or not .

Examine the values of x that is from -3 to 3, the f(x) values are both positive and negative . hence f(x)>0 is false over the interval (-∞,3)

Looking at the the interval from 0 to 2, the f(x) values are 0 and negative. Hence, f(x)<=0 over the interval [0,2]

When you look over the interval (-1,1), the f(x) values are said to be both positive and negative and as such, f(x)<0 is false over the interval (-1,1)

When you look at the interval (-2,0) , the f(x)  is positive and as such, f(x)>0 over the interval (-2,0)

Looking at the interval  [2,∞), f(x) is positive and as such, f(x)>=0 over the interval [2,∞)

Therefore, Option 2, 4 and 5 are correct.

Learn more about interval from

brainly.com/question/14454639

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Step-by-step explanation:

Hope this helps

8 0
3 years ago
Read 2 more answers
Suppose that c varies jointly with d and the square of g, and c = 30 when d = 15 and g = 2.
maksim [4K]

Answer:

d = \dfrac{3}{16}

Step-by-step explanation:

c varies jointly with d and the square of g

c\propto dg^2

c=kdg^2

where, k is constant of proportionality.

Put the given value c = 30 when d = 15 and g = 2 and find out k

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k=\dfrac{1}{2}

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If c = 6 and g = 8 then d = ?

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Hence, The value of d is \dfrac{3}{16}

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