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3 years ago

Evaluate and show work 1/2 divided by 36 + 16 times 1/72

Mathematics

2 answers:

melamori03 [73]3 years ago

7 0

Answer:

1/2

Step-by-step explanation:

1/2 times 52, 36 + 16, would be 26. Then I multiplied by 1/72 to get 1/2

Gwar [14]3 years ago

6 0

Answer:

17/72

Step-by-step explanation:

Let's make this equation easier to read.

.5/36 + 16(1/72)

Start by remembering PEMDAS

Parentheses

Exponents

Multiply

Divide

Add

Subtract

This is the order in which you will solve the problem. So multiplication first, then division, and finally addition.

Multiply 16 and 1/72

16/1 x 1/72 multiply across to get 16/72 which can be reduced to 2/9.

Divide .5 (or 1/2) by 36. To do this, multiply by the reciprocal of 36

1/2 x 1/36 = 1/72

Your equation is now

2/9 + 1/72

Your common denominator is 72, so multiply top and bottom of 2/9 by 8.

16/72 + 1/72 = 17/72

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3 years ago

Translate (11, 3) 8 units up and 14 units left<br> What’s the answer

Lubov Fominskaja [6]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Determine what type of model best fits the given situation:

lyudmila [28]

Let value intially be = P

Then it is decreased by 20 %.

So 20% of P = $\frac{20}{100} \times P = 0.2P$

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P = $\frac{20}{100} \times 0.8P = (0.2)(0.8P)$

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

$=P(0.8)^{2}$ -------------------------(2)

Similarly after 3 years, this value $P(0.8)^{2}$ will again be decreased by 20 %

so 20% of $P(0.8)^{2} \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}$

So after 3 years value is decreased by $(0.2)P(0.8)^{2}$

so value after 3 years will be = $P(0.8)^{2} - (0.2)P(0.8)^{2}$

taking $P(0.8)^{2}$ common out we get $P(0.8)^{2}(1-0.2)$

$P(0.8)^{2}(0.8)$

$P(0.8)^{3}$ -----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or $P(0.8)^{1}$

value after 2 years is $P(0.8)^{2}$

value after 3 years is $P(0.8)^{3}$

so value after x years will be $P(0.8)^{x}$ ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

$y = P(0.8)^{x}$ where y is the value after x years

so thats the final answer

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Answer:

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Step-by-step explanation:

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