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RideAnS [48]
3 years ago
15

Which is equivalent to RootIndex 3 StartRoot 8 EndRoot Superscript x?

Mathematics
1 answer:
sergeinik [125]3 years ago
8 0

Solving  RootIndex 3 StartRoot 8 EndRoot Superscript x we get =2^x

Step-by-step explanation:

We need to find equivalent to RootIndex 3 StartRoot 8 EndRoot Superscript x

Writing in mathematical form:

(\sqrt[3]{8})^x

Solving:

We know 8= 2x2x2= 2^3

and \sqrt[3]{x}=x^{\frac{1}{3}}

Applying these rules:

=((2^3)^{\frac{1}{3}})^x

=(2^{\frac{3}{3}})^x\\

=2^x

So, solving  RootIndex 3 StartRoot 8 EndRoot Superscript x we get =2^x

Keywords: Radical Expression

Learn more about Radical Expression at:

  • brainly.com/question/7153188
  • brainly.com/question/10534381

#learnwithBrainly

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The measure of the first angle in a right triangle is 10 less than the measure of the second angle.The third angle is the right
prohojiy [21]
Triangle angles add up to a total of 180 degrees.

X will substitute for the second angle.

(X - 10) will substitute for the first angle, which is 10 less than the second angle (x).

The third angle is a right angle which is 90 degrees.

The equation therefore, the summation of the angles:
90 + x + (x - 10) = 180

Join like terms:
80 + 2x = 180

Solve for x:
2x = 100
x = 50

This means that:
The second angle is 50.
The first angle is 50 - 10 which is 40.

We can double check by adding the angles 50, 40, and 90 together to get 180.
5 0
3 years ago
Can some one awser some of those please thank you please :)
Alexeev081 [22]
5. 1,000,000
6. 9,300
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8. 5, 280
9. 810
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12. 76, 000

Those are a few answers for you. Try looking back at your notes (If you copied any) to figure out your problems :)
4 0
3 years ago
Read 2 more answers
Is the number ���12/4 rational
zalisa [80]
Yes because 

12/4 is 3
8 0
3 years ago
A table is shaped like a regular hexagon. The length of one side is (7x + 4) in. and the length of another side is (8x − 1) in.
kirza4 [7]

Answer:

39 in.

Step-by-step explanation:

As  it's a regular hexagon all sides are congruent.

Therefore

7x + 4 = 8x - 1

4 + 1 = 8x - 7x

x = 5.

So the required length is 7(5) + 4 = 39 in.

8 0
2 years ago
Some people think it is unlucky if the 13th day of month falls on a Friday. show that in that there year (non-leap or leap) ther
Vlad1618 [11]
<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths. Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday. Jan - Friday Feb - Monday Mar - Monday Apr - Thursday May - Saturday Jun - Tuesday Jul - Thursday Aug - Sunday Sep - Wednesday Oct - Friday Nov - Monday Dec - Wednesday Now let's count how many times for each weekday, the 13th falls there. Sunday - 1 Monday - 3 Tuesday - 1 Wednesday - 2 Thursday - 2 Friday - 2 Saturday - 1 The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc. So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get: Jan - Friday Feb - Monday Mar - Tuesday Apr - Friday May - Sunday Jun - Wednesday Jul - Friday Aug - Monday Sep - Thursday Oct - Saturday Nov - Tuesday Dec - Thursday And the weekday totals are: Sunday - 1 Monday - 2 Tuesday - 2 Wednesday - 1 Thursday - 2 Friday - 3 Saturday - 1 And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year. And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
5 0
3 years ago
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