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masya89 [10]
4 years ago
7

Can you please help me simplify the expression?

Mathematics
2 answers:
Ugo [173]4 years ago
8 0

Answer:

-5h - 2

Step-by-step explanation:

3h - 2(1+4h)

Let's first get rid of the ( ) by performing the multiplication:

3h - 2*1 + -2*4h = 3h - 2 - 8h

Now we can subtract 8h from 3h, and that's as simple as it gets:

3h - 2 - 8h = -5h - 2

Harrizon [31]4 years ago
7 0

-5h-2 is the correct answer.

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If a^2+1/a^2=79 where (a>0), find the value of a^3 +1/a^3
Nastasia [14]

Note the binomial expansion,

(<em>a</em> + 1/<em>a</em>)³ = <em>a</em> ³ + 3<em>a</em> + 3/<em>a</em> + 1/<em>a</em> ³

so

<em>a</em> ³ + 1/<em>a</em> ³ = (<em>a</em> + 1/<em>a</em>)³ - 3 (<em>a</em> + 1/<em>a</em>)

Similarly,

(<em>a</em> + 1/<em>a</em>)² = <em>a</em> ² + 2 + 1/<em>a</em> ²

We're given <em>a</em> ² + 1/<em>a</em> ² = 79, so

(<em>a</em> + 1/<em>a</em>)² - 2 = 79

(<em>a</em> + 1/<em>a</em>)² = 81

<em>a</em> + 1/<em>a</em> = ±9

but <em>a</em> > 0, so we ignore the negative solution.

Then

<em>a</em> ³ + 1/<em>a</em> ³ = 9³ - 3×9 = 702

7 0
3 years ago
An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
sladkih [1.3K]

Answer:

(a) The probability that a randomly selected adult is either overweight or obese is 0.569.

(b) The probability that a randomly selected adult is neither overweight nor obese is 0.431.

Step-by-step explanation:

Let <em>A</em> = a person is over weight and <em>B</em> = a person is obese.

The information provided is:

An adult is considered overweight if the BMI ≥ 25 but BMI < 30.

An obese adult will have a BMI ≥ 30.

According to the range of BMI the events A and B are independent.

P (A) = 0.331 and P (B) = 0.357.

(a)

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

               = P (A) + P (B) - P (A)×P (B)

               =0.331+0.357-(0.331\times0.357)\\=0.569833\\=0.569

Thus, the probability that a randomly selected adult is either overweight or obese is 0.569.

(b)

Compute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(A^{c}\cup B^{c})=1-P(A\cup B)=1-0.569=0.431

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.431.

4 0
4 years ago
The length of a rectangular piece of cardboard is 1 4/5 meters. If its width is 2/3 of its length, what is the width of the card
Lynna [10]

Answer:

6/5 or 1.2

Step-by-step explanation:

First, you need to multiply the length by how much the width is of the length, which is 2/3, so you multiply 1 4/5 by 2/3 and you get 6/5. If you want the decimal form, then the answer will be 1.2.

7 0
3 years ago
alexa and rosa read books that have the same number of pages.alexa book is divided into 8 equal chapters. rosa book is divided i
Molodets [167]
Alexa read 3/8 and rosa read 3/6
8 0
3 years ago
Read 2 more answers
The 7 in 507.3 has ten times the value of the 7 in?<br> Is it 12.007, 725, 67 or 30.76.
defon
I believe the answer would be 30.76
4 0
3 years ago
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