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Viefleur [7K]
3 years ago
10

To the nearest thousand the population of Texas was estimated to be 24,327,000 in 2008. Describe the actual population that Texa

s could have had in 2008
Mathematics
1 answer:
azamat3 years ago
5 0
To the nearest thousand, population is 24,327,000

that means the population could range between 24,326,500 and 24,327,499
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Find the domain of y= 2/3 square root of x+5-4
3241004551 [841]
Set the radicand in
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Set-Builder Notation:
{
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(
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[
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Set-Builder Notation:
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Determine the domain and range.
4 0
2 years ago
How do you solve capacity?​
ArbitrLikvidat [17]

Answer:

Cube and rectangle- length times with times height

5 0
3 years ago
Va rog e urgent am nevoie de raspuns
Lady_Fox [76]

Answer: 4

<u>Explanation:</u>

f(x) = 2x - 1

f(√2) = 2√2 - 1

f(1) = 2(1) - 1

     =  2 - 1

     =     1

f(√3)  = 2√3 - 1

*******************************************************

\frac{f(\sqrt{2})-f(1)}{\sqrt{2}-1} +\frac{f(\sqrt{3})-f(\sqrt{2})}{\sqrt{3}-\sqrt{2}}

= \frac{(2\sqrt{2}-1)-1}{\sqrt{2}-1} +\frac{(2\sqrt{3}-1)-(2\sqrt{2}-1)}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1} +\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1}(\frac{\sqrt{2}+1}{\sqrt{2}+1})+\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}})

= \frac{4+2\sqrt{2}-2\sqrt{2}-2}{2 - 1} + \frac{6 +2\sqrt{6}-2\sqrt{6}-4}{3-2}

= \frac{2}{1} +\frac{2}{1}

= 4

3 0
2 years ago
Use
Brums [2.3K]

Answer:

I don't know what inductive reasoning is but, the next two numbers in the pattern are -243 and +729

Step-by-step explanation:

each number is being multiplied by a factor of (-3).

8 0
2 years ago
Sally buys a hang bag. There was a discaount of 35percent . If Sally paid $56.80 what was the original price?
Ksju [112]
Discount = 35%
Percentage after discount = 100% - 35% = 65%

65% = $56.80
1% = $0.8738
100% = $87.38

Answer: The original price is $87.38
4 0
3 years ago
Read 2 more answers
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