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Phoenix [80]
3 years ago
9

1. Write the standard form of the 6 in the numbel . * 1 p 792,348,612

Mathematics
1 answer:
madam [21]3 years ago
7 0

Answer:

600

Step-by-step explanation:

5&---------------------

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It costs $56 to rent a tent for 7 days. Write an equation that represents the cost to rent a camping tent for xdays.
Lapatulllka [165]
the equation might be 8x=y.
7 0
3 years ago
Is 5.3 repeating rational
Goshia [24]

Repeating decimals are considered rational numbers because they can be represented as a ratio of two integers. If a number is terminating or repeating, it must be rational if a decimal is both non terminating and non repeating, the number is irrational.

So yes.

4 0
3 years ago
(0,65-3,21)por(-8,12)=?
PIT_PIT [208]

Answer:  =20,7872

Step-by-step explanation:

For this exercise it is important to remember the multiplication of signs. Notice that:

(+)(+)=+\\\\(-)(-)=+\\\\(-)(+)=-\\\\(+)(-)=-

In this case you have the following expression given in the exercise:

(0,65-3,21)(-8,12)

Then you can follow the steps shown below in order to solve it:

Step 1: You must solve the subtraction of the numbers 0,65 and 3,21. Then:

=(-2,56)(-8,12)

Step 2: Now you must find the product of the decimal numbers above. In order to do that you must multiply the numbers.

(As you can notice, both are negative, therefore you know that the product will be positive).

Then, you get that the result is the following:

 =20,7872

5 0
3 years ago
2 4 8 16 what is the rule and the sixth number in the sequence answers
aleksklad [387]
The sequence is incrementing by
{2}^{n}
Therefore, the sixth number is
{2}^{6}
Which is 64.
7 0
3 years ago
Show that the equation x^3+6x-5=0 has a solution between x=0 and x=1
Mnenie [13.5K]

Answer:

Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.

Step-by-step explanation:

The equation f(x) given is:

f(x) = x^3+6x-5

For x = 0. the value of the expression is:

f(0) = 0^3+0-5\\f(0) = -5

For x = 1, the value of the expression is:

f(1) = 1^3+6-5\\f(1)=2

Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.

In other words, there is at least one solution for the equation between x=0 and x=1.

6 0
3 years ago
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