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nataly862011 [7]
3 years ago
6

There are 200 marbles in the bag. John picks 10 marbles out of

Mathematics
1 answer:
aleksklad [387]3 years ago
3 0

Answer:

C. 60 blue marbles

Step-by-step explanation:

10 marbles are taken out of the bag.                                    

6 marbles taken out are red.

So the other 4 marbles taken out are blue.

200 ÷ 4 = 50

and 50 is close to 60.

Let me know if this helps!

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7(x-2)=28 Can somebody help me and show the steps?
olya-2409 [2.1K]
Since the 7 is in front of the parentheses you must multiply it by everything inside the parentheses. 7 times X is 7X and 7 times -2 is -14. Then you need to make the X be by itself on one side, to do that you have to add 14 to -14 to make it disappear, and also add 14 to the other side, to 28. Then you divide both sides by 7 to make X be by itself.

7(x-2)=28
7X-14=28
7x=42
X=6

6 0
2 years ago
What is a 33333333...2222
pshichka [43]
Are you sure that is a real question

6 0
3 years ago
NEED HELP ASAP ILL MARK BRAINLIEST IF ITS RIGHT!!!​
Leto [7]

Answer:

f(1)=-11

f(n)=f(n-1)+22

Step-by-step explanation:

Base equation:

f(n)=-11+22(n-1)

f(1)=-11+22(1-1)

f(1)=-11+0

f(1)=-11

f(n)=-33+22n

f(n-1)=-11+22(n-2)

f(n-1)=-55+22n

-33+22n=-55+22n+x

22n gets canceled out

-33=-55+x

-33+55=x

x=22

f(n)=f(n-1)+22

5 0
3 years ago
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prohojiy [21]

Answer:

is the root of 5 not equal why

8 0
3 years ago
Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
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