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disa [49]
3 years ago
9

A piece of machinery depreciates $5000 the first year,

Mathematics
1 answer:
eimsori [14]3 years ago
4 0

Answer:

A- $3500

Step-by-step explanation:

The depreciation rate is $300

$5000, $4700, $4400

$300 x 3 = 900

$4400 - 900 =

$3500.

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PLEASE HELP ME WITH THIS PROBLEM IVE BEEN ON IT FOR 3 DAYS NOW!
Len [333]
85 x 3 = 255
 first 2 test grades were 81 and 86:
81 + 86 = 167

Need to find the 3rd to have exact average 85:
255 - 167 = 88

So her 3rd test needs to make 88 to make average of 85

Double check: 81 + 86 + 88 =  255 = 85 + 85 + 85
3 0
3 years ago
4. Is there a variable factor common to 24a and 6?​
Cloud [144]

Answer:

There is no variable factor common to 24a and 6

Step-by-step explanation:

First number = 24a

Other number = 6

Pime factorisation of 24a = 2×2×3×2×a

Pime factorisation of 6= 3×2

Common factors = 6

Hence, there is no variable factor common to 24a and 6, Other common factor is 6.

7 0
3 years ago
Lily quinn makes $12.50 and hour. she works four hours on monday, six hours on tuesday, nine hours on wednesday, three hours on
AnnZ [28]
The answer is A. Her gross pay is $362.50.
4 0
3 years ago
If P and S are rational numbers and R and Q are irrational numbers, which of these statements is true?
masha68 [24]

Answer: B. The product of P and S is always a rational number.

Step-by-step explanation: The product AND sum of two rational numbers will ALWAYS be rational.

8 0
3 years ago
If k is an integer, what is <br> k + 1<br> /2<br> !<br> ? Why?
Amanda [17]

If k\in\mathbb{Z} than k+1\in\mathbb{Z} however any \mathbb{Z} divided by 2 is in \mathbb{Q}.

But it is not so simple.

If k is odd then k+1 is even but when even number is divided by 2, you get an odd number which is in \mathbb{Z}.

However if k is even then k+1 is odd and when we divide that by 2, you get a number like for example \frac{5}{2} which is not in \mathbb{Z} but rather inside \mathbb{Q}. There are fortunately no irrational numbers.

There is also a problem with zero, zero is neither odd nor even but still an integer. If your k happens to be -1, k+1=0, and zero divided by 2 is still just 0 which is in \mathbb{Z}.

So to sum up,

If k is odd and not -1, then \frac{k+1}{2}\in\mathbb{Z}.

Or to put it in math \forall k\in\mathrm{odd}-\{-1\}\cup\{0\}\implies\frac{k+1}{2}\in\mathbb{Z}.

If k is even or -1, then \frac{k+1}{2}\in\mathbb{Q}.

Or to put it in math \forall k\in\mathrm{even}\cup \{-1\}\implies\frac{k+1}{2}\in\mathbb{Q}.

<u>Notes</u>:

\mathbb{Z} is a set of integers.

\mathbb{Q} is a set of rational numbers.

\mathbb{Z}\subset\mathbb{Q} means integers are a subset of rational numbers, that is, every integer is a also a fraction, however not every fraction is an integer.

Hope this helps :)

8 0
3 years ago
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