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ipn [44]
3 years ago
13

Nick starts with 20 milligrams of a radioactive substance. The amount of the substance decreases by 14 each week for a number of

weeks, w. He writes the expression 20(14)w to find the amount of radioactive substance remaining after w weeks. Autumn starts with 5 milligrams of a radioactive substance. The amount of the substance decreases by 50% each week for a number of weeks, w. She writes the expression (1?0.5)w to find the amount of radioactive substance remaining after w weeks. Use the drop-down menus to explain what each part of Nick's and Autumn's expressions mean.
Mathematics
2 answers:
zlopas [31]3 years ago
8 0

Answer:

Eric's expression is :

20(\frac{1}{4})^w

and Andrea's is :

(1-0.5)^w

In Eric's expression, 20 represents the initial amount of substance with which he has started the experiment. \thinspace \frac{1}{4}\thinspace is the amount of substance left after each time period (in this case, each week).The variable w in this case represents the number of weeks.

Andrea's expression can be written as :

1\cdot (1-0.5)^w

The one outside of parentheses represents the initial amount of the substance. The one inside of parentheses represents 100% of the original amount of the substance. 0.5 represents the 50% of the substance that is lost each time period. The variable w in this case represents the number of weeks.

uranmaximum [27]3 years ago
4 0

Answer:

Step-by-step explanation:

Nick starts with 20 milligrams of a radioactive substance. The amount of the substance decreases by 12 each week for a number of weeks, w. He writes the expression 20(12)w to find the amount of radioactive substance remaining after w weeks.

Autumn starts with 1 milligram of a radioactive substance. The amount of the substance decreases by 50% each week for a number of weeks, w. She writes the expression (1−0.5)w to find the amount of radioactive substance remaining after w weeks.

Use the drop-down menus to explain what each part of Nick's and Autumn's expressions mean.

CLEAR CHECK

Nick's Expression: 20(12)w

12:  

w:  

20:  

(12)w:  

Autumn's Expression: (1−0.5)w

w:  

0.5:  

1−0.5:

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lakkis [162]

Answer:

Numerator = 2(b^2+a^2)    or equivalently 2b^2+2a^2

Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3

Step-by-step explanation:

Let

S = 2b/(b+a)^2 + 2a/(b^2-a^2)   factor denominator

= 2b/(b+a)^2 + 2a/((b+a)(b-a))     factor denominators

= 1/(b+a) ( 2b/(b+a) + 2a/(b-a))    find common denominator

= 1/(b+a) ((2b*(b-a) + 2a*(b+a))/((b+a)(b-a))   expand

= 1/(b+a)(2b^2-2ab+2ab+2a^2)/((b+a)(b-a))  simplify & factor

= 2/(b+a)(b^2+a^2)/((b+a)(b-a))  simplify & rearrange

= 2(b^2+a^2)/((b+a)^2(b-a))

Numerator = 2(b^2+a^2)    or equivalently 2b^2+2a^2

Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3

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3 years ago
Is (x-3) a factor of f(x) = x^2 - 32?
Leokris [45]

Answer:

No

Explanation:

Here, we want to use the factor theorem to check if the given linear expression is a factor of the binomial

Now, according to the factor theorem, a factor of a polynomial would leave no remainder when divided by it

Mathematically, it means when we substitute the factor value into the polynomial, it is expected that the remainder is zero is the substituted is a factor of the polynomial

We set x-2 to zero:

\begin{gathered} \text{ x-2 = 0} \\ x\text{ = 2} \end{gathered}

Now, we substitute 2 into the polynomial as follows:

f(2)\text{ = }2^2-32\text{ = 4-32 = -28}

There is a remainder of -28 and thus, the linear factor is not a factor of the binomial

3 0
1 year ago
The photography club decided to go on a field trip to the Andy Warhol museum. They had to break into smaller groups once they go
Diano4ka-milaya [45]
A)
Let x represent the cost of 1 student, and y the cost of 1 teacher.

B)
In the first group, there's 25 students and 2 teachers. Their total cost is $97.50
So 25x + 2y = 97.50
In the second group, there's 32 students and 3 teachers. Their total cost is $127
So 32x + 3y = 127

We get the following system of equations:
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)

C)
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)

In equation (1)
25x + 2y = 97.50
25x + 2y - 2y = 97.50 - 2y
25x = 97.50 - 2y
25x / 25 = 97.50/25 - 2y/25
x = 3.9 - (2/25)y

In equation (2), let's replace x by its algebraic value
32x + 3y = 127
32(-2/25y + 3.9) + 3y = 127
11/25y + 124.8 = 127
11/25y + 124.8 - 124.8 = 127 - 124.8
11/25y = 2.2
(11/25y) / (11/25) = 2.2 / (11/25)
y = 5

x = -2/25y + 3.9
x = -2/25 * 5 + 3.9
x = 3.5

So the cost of each student is $3.5, and the cost of each teacher is $5.

Hope this helps! :)




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