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Umnica [9.8K]
1 year ago
11

Alison is buying binders for school. Small binders cost $3 each, and large binders cost $5 each. If Alison needs to buy at least

12 binders and has no more than $45 to spend, what is the maximum number of large binders she can buy?
a. 4

b 5

c 8

d 9
Mathematics
2 answers:
prisoha [69]1 year ago
8 0

Answer:

4 large folders     ( and 8 small folders....with $ 1 left over)

Step-by-step explanation:

x = small    y = large folders

x+y >= 12      x >=<u> 12-y </u>

3x + 5y <= 45

3 (12-y) + 5y <=45

36 -3y + 5y <= 45

    y <= 4.5               you cannot buy .5 of a large folder, so <u>4 large folders</u>

Mariulka [41]1 year ago
4 0

Answer:

9

Step-by-step explanation:

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Factor completely x2 + 25
Ganezh [65]

Answer:

(x - 5i)(x + 5i)

Step-by-step explanation:

Write this as x^2 + 25.  Solving for x^2, we get x^2 = -25.

Recall that the square root of a negative number is an imaginary one.  √(-25) equals ±5i (there are two roots, one positive and one negative).

Then x = ±5i.

In factored form, the given expression is (x - 5i)(x + 5i).

6 0
3 years ago
What type of number is - 36/12
Vitek1552 [10]

Answer:

It is an improper fraction.

Step-by-step explanation:

With fractions, if the numerator is equal to or grater than the denominator, the fraction is considered an improper fraction.

6 0
2 years ago
You have the first season of your favorite television show on a set of DVDs. The set
Misha Larkins [42]

Answer: There are 45 combinations.

Step-by-step explanation:

If the order in which you watch the episodes does not matter, then we can use the combinatory defined as:

If we have N elements, and we want to make groups of K elements, we have

C(N,K) = \frac{N!}{(N-K)!*K!}

combinations.

In this case, we have a total of N = 10 episodes, and we want to make groups of  K = 2.

Then we have:

C = (10, 2) = \frac{10!}{8!*2!} = (10*9)/(2) = 5*9 = 45

There are 45 combinations.

8 0
3 years ago
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Can you provide a picture or what is going on in the picture and or text?

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3 years ago
How derivative of this absolute value function is like this!
Daniel [21]

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|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}

When x,

|x| = x \implies \dfrac{d|x|}{dx} = 1

When x,

|x| = -x \implies \dfrac{d|x|}{dx} = -1

The derivative does not exist at x=0, since the one-sided limits

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and

\displaystyle \lim_{x\to0^+} f'(x) = +1

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So the derivative of |x| is

\dfrac{d|x|}{dx} = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \\ \text{undefined} & \text{if } x = 0\end{cases}

Now we can write this as

\dfrac{d|x|}{dx} = \dfrac x{|x|} = \dfrac{|x|}x

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x > 0 \implies |x| = x \implies \dfrac{|x|}x = \dfrac xx = 1

and

x < 0 \implies |x| = -x \implies \dfrac{|x|}x = -\dfrac xx = -1

6 0
1 year ago
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