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a_sh-v [17]
3 years ago
5

What is the following simplified product? Assume x greater-than-or-equal-to 0 2 StartRoot 8 x cubed EndRoot (3 StartRoot 10 x Su

perscript 4 Baseline EndRoot minus x StartRoot 5 x squared EndRoot) 24 x cubed StartRoot 5 x EndRoot minus 4 x squared StartRoot 10 x EndRoot 24 x cubed StartRoot 5 x EndRoot minus 4 x cubed StartRoot 10 x EndRoot 24 x cubed StartRoot 5 EndRoot minus 4 x cubed StartRoot 10 x EndRoot 24 x cubed StartRoot 5 x EndRoot + 4 x cubed StartRoot 10 x EndRoot
Mathematics
2 answers:
Lady bird [3.3K]3 years ago
5 0
Wow that’s a lot let me try to help
Sedbober [7]3 years ago
4 0

Answer:

The answer to this should be D on ed2020

Step-by-step explanation:

There's another post for this question where Cheerchar13 had answered.

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What is the solution to 3 x 2. 4 greater-than-or-equal-to 3. 0? x greater-than-or-equal-to 0. 2 x greater-than-or-equal-to 0. 6
geniusboy [140]

Answer:

Excuse me if this is wrong but I think you have to multiply 3 x 2.4 which is 7.2 and that is greater that 3.0.

Step-by-step explanation:

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3 years ago
Your classmate is unsure about how to use side lengths
DENIUS [597]

Answer:

you look at the lengths some triagles have small parallel lines anè or two of them to determine if they are equal or different from one another

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3 years ago
Girardo is using the model below to solve the equation 3x+1=4x+(-4)
V125BC [204]

Answer:

x=5

Step-by-step explanation:

3x+1= 4x+(-4)

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-1x= -5 divided by -1

x= 5

5 0
3 years ago
Passes through (8,7) and (0,0)
quester [9]

Answer:

6,3

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
A worker was paid a salary of $10,500 in 1985. Each year, a salary increase of 6% of the previous year's salary was awarded. How
Mazyrski [523]
Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.

Here is how much the worker earned each year:


In the year 1985 the worker earned <span>$10,500. 

</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:

                                            $10,500(1+0.06).



In the year 1987 the worker earned

$10,500(1+0.06) + 0.06[$10,500(1+0.06)].

Now we can factorize $10,500(1+0.06) and write the earnings as:

$10,500(1+0.06) [1+0.06]=$10,500(1.06)^2.


Similarly we can check that in the year 1987 the worker earned $10,500(1.06)^3, which makes the pattern clear. 


We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:

10,500+10,500(1.06)^1+10,500(1.06)^2+10,500(1.06)^3...10,500(1.06)^{26}.

Factorizing, we have

=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]

We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula

\sum_{i=1}^{n} a_i= a(\frac{1-r^n}{1-r}) (where a is the first term and r is the common ratio) we have:

\sum_{i=1}^{26} a_i= 1(\frac{1-(1.06)^{26}}{1-1.06})= \frac{1-4.55}{-0.06}= 59.17.



Finally, multiplying 10,500 by 59.17 we have 621.285 ($).


The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating (1.06)^{26}.


Answer: D



4 0
3 years ago
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