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vovikov84 [41]
3 years ago
10

Find two solutions to the equation (x^3 − 64)(x^5 − 1) = 0.

Mathematics
1 answer:
Rashid [163]3 years ago
4 0

Answer:

the two roots are x = 1 and x = 4

Step-by-step explanation:

Data provided in the question:

(x³ − 64) (x⁵ − 1) = 0.

Now,

for the above relation to be true the  following condition must be followed:

Either  (x³ − 64) = 0 ............(1)

or

(x⁵ − 1) = 0 ..........(2)

Therefore,

considering the first equation, we have

(x³ − 64) = 0

adding 64 both sides, we get

x³ − 64 + 64 = 0 + 64

or

x³ = 64

taking the cube root both the sides, we have

∛x³ = ∛64

or

x = ∛(4 × 4 × 4)

or

x = 4

similarly considering the equation (2) , we have

(x⁵ − 1) = 0

adding the number 1 both the sides, we get

x⁵ − 1 + 1 = 0 + 1

or

x⁵ = 1

taking the fifth root both the sides, we get

\sqrt[5]{x^5}=\sqrt[5]{1}

also,

1 can be written as 1⁵

therefore,

\sqrt[5]{x^5}=\sqrt[5]{1^5}

or

x = 1

Hence,

the two roots are x = 1 and x = 4

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The space shuttle travels at 27 870 km per hour.
torisob [31]

Answer:

B

Step-by-step explanation:

3 x 27870 = 83610

4 0
3 years ago
I don't understand these problems
uysha [10]
So given the values on the number line, AB and BC, adding them together will get the length of AC. If AB equals x-3 and BC equals 5, the value of AC should be x-3+5. Since you are given the length of AC, 21, you can create an equation to solve for x. This equation would be x-3+5=21. Isolating the x on the left (or right hand side) gets the value of x to be 19. The same thing goes for the 2nd question.
6 0
3 years ago
I need help solving this step by step. Work would be nice so I can look and understand it so I can try a different problem. ​
zimovet [89]

Answer:

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Step-by-step explanation:

3 0
3 years ago
Let N be the smallest positive integer whose sum of its digits is 2021. What is the sum of the digits of N + 2021?
kondor19780726 [428]

Answer:

10.

Step-by-step explanation:

See below for a proof of why all but the first digit of this N must be "9".

Taking that lemma as a fact, assume that there are x digits in N after the first digit, \text{A}:

N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{$x$ digits}}}, where x is a positive integer.

Sum of these digits:

\text{A} + 9\, x= 2021.

Since \text{A} is a digit, it must be an integer between 0 and 9. The only possible value that would ensure \text{A} + 9\, x= 2021 is \text{A} = 5 and x = 224.

Therefore:

N = \overline{5 \, \underbrace{9 \cdots 9}_{\text{$224$ digits}}}.

N + 1 = \overline{6 \, \underbrace{000 \cdots 000000}_{\text{$224$ digits}}}.

N + 2021 = 2020 + (N + 1) = \overline{6 \, \underbrace{000 \cdots 002020}_{\text{$224$ digits}}}.

Hence, the sum of the digits of (N + 2021) would be 6 + 2 + 2 = 10.

Lemma: all digits of this N other than the first digit must be "9".

Proof:

The question assumes that N\! is the smallest positive integer whose sum of digits is 2021. Assume by contradiction that the claim is not true, such that at least one of the non-leading digits of N is not "9".

For example: N = \overline{(\text{A})\cdots (\text{P})(\text{B}) \cdots (\text{C})}, where \text{A}, \text{P}, \text{B}, and \text{C} are digits. (It is easy to show that N contains at least 5 digits.) Assume that \text{B} \! is one of the non-leading non-"9" digits.

Either of the following must be true:

  • \text{P}, the digit in front of \text{B} is a "0", or
  • \text{P}, the digit in front of \text{B} is not a "0".

If \text{P}, the digit in front of \text{B}, is a "0", then let N^{\prime} be N with that "0\!" digit deleted: N^{\prime} :=\overline{(\text{A})\cdots (\text{B}) \cdots (\text{C})}.

The digits of N^{\prime} would still add up to 2021:

\begin{aligned}& \text{A} + \cdots + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + 0 + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}.

However, with one fewer digit, N^{\prime} < N. This observation would contradict the assumption that N\! is the smallest positive integer whose digits add up to 2021\!.

On the other hand, if \text{P}, the digit in front of \text{B}, is not "0", then (\text{P} - 1) would still be a digit.

Since \text{B} is not the digit 9, (\text{B} + 1) would also be a digit.

let N^{\prime} be N with digit \text{P} replaced with (\text{P} - 1), and \text{B} replaced with (\text{B} + 1): N^{\prime} :=\overline{(\text{A})\cdots (\text{P}-1) \, (\text{B} + 1) \cdots (\text{C})}.

The digits of N^{\prime} would still add up to 2021:

\begin{aligned}& \text{A} + \cdots + (\text{P} - 1) + (\text{B} + 1) + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}.

However, with a smaller digit in place of \text{P}, N^{\prime} < N. This observation would also contradict the assumption that N\! is the smallest positive integer whose digits add up to 2021\!.

Either way, there would be a contradiction. Hence, the claim is verified: all digits of this N other than the first digit must be "9".

Therefore, N would be in the form: N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{many digits}}}, where \text{A}, the leading digit, could also be 9.

6 0
3 years ago
Don has a bag that weighs 2 oz, and Jamie has a bag that weighs 1 oz. Don put some marbles in his bag each weighing 9.4 oz, and
Andrej [43]
total weight of both bags = 3 oz + 20.6m oz

Don's bag = 2 oz
Jamie's bag = 1 oz
Don's marbles = 9.4 oz * m
Jamie's marlbes = 11.2 oz * m

Total weight of Don's bag = 2 oz + 9.4m oz
Total weight of Jamie's bag = 1 oz + 11.2m oz

2 oz + 9.4m oz + 1 oz + 11.2m oz = 3 oz + 20.6m oz
6 0
3 years ago
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