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3 years ago

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.

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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.

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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 k + 1 /2 ! ? 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 $\forall k\in\mathrm{even}\cup \{-1\}\implies\frac{k+1}{2}\in\mathbb{Q}$ .

Notes:

$\mathbb{Z}$ 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