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White raven [17]
2 years ago
14

Write 44 as a product of prime factors.

Mathematics
1 answer:
Ann [662]2 years ago
3 0

<u>Answer:</u>

  • Hence, 11 x 2 x 2 is the product of prime factors.

<u>Step-by-step explanation:</u>

<u>Let's divide by 2's.</u>

  • => 44 = 22 x 2
  • => 22 x 2 = 11 x 2 x 2

<em>Since 11 is a prime number, we cannot simplify it anymore. Hence, 11 x 2 x 2 is the product of prime factors.</em>

<em />BrainiacUser1357<em />

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Answer:

5 5/8

Step-by-step explanation:

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3 years ago
Which statement describes if there is an extraneous solution to the equation √x-3 = x-5? A. there are no solutions to the equati
Brrunno [24]
Remember that <span>an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.
 
Lets solve our equation to find out what is the extraneous solution:
</span>\sqrt{x-3} =x-5
(\sqrt{x-3})^2 =(x-5)^2
x-3=x^2-10x+25
x^2-11x+28=0
(x-4)(x-7)=0
x-4=0 and x-7=0
x=4 and x=7
<span>
So, the solutions of our equation are </span>x=4 and x=7. Lets replace each solution in our original equation to check if they are valid solutions:
- For x=7
\sqrt{x-3} =x-5
\sqrt{7-3} =7-5
\sqrt{4} =2
2=2
We can conclude that 7 is a valid solution of the equation.

- For x=4
\sqrt{x-3} =x-5
\sqrt{4-3} =4-5
\sqrt{1} =1
1 \neq 1
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: <span>D. the extraneous solution is x = 4</span>
7 0
3 years ago
8. Which function is shown on the graph below?
Dafna11 [192]

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Step-by-step explanation:

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6 0
2 years ago
Find the equivalent expression using the same bases (97*59)^9
irina [24]

Answer:

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Decimal form : 6.58587848 • 10^33

Step-by-step explanation: Evaluate.

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4 0
3 years ago
Travis jogged to his friend's house 10 miles away and then got a ride back home. It took him 2 hours longer to jog there than ri
spin [16.1K]

Using the relation between velocity, distance and time, it is found that his jogging rate was of 4.5 mph.

<h3>What is the relation between velocity, distance and time?</h3>

Velocity is <u>distance divided by time,</u> hence:

v = \frac{d}{t}

Jogging, his velocity was of v, while the time was of t + 2, for a distance of 10 miles, hence:

v = \frac{10}{t + 2}

Riding, his velocity was of 16, while the time was of t, also for a distance of 10 miles, hence:

v + 16 = \frac{10}{t}

Then, considering the first equation and replacing in the second:

\frac{10}{t + 2} + 16 = \frac{10}{t}

\frac{10}{t} - \frac{10}{t + 2} = 16

\frac{10t + 20 - 10t}{t(t + 2)} = 16

16t² + 32t - 20 = 0 -> t = 0.5.

Then his jogging rate in mph was:

v = \frac{10}{0.5 + 2} = 4.5

More can be learned about the relation between velocity, distance and time at brainly.com/question/24316569

#SPJ1

8 0
2 years ago
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