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2 years ago

You went shopping for back to school and bought packs of pens and pencils. a pack of

Mathematics

1 answer:

Scrat [10]2 years ago

7 0

Answer:

5 packs of pens and 8 packs of pencils.

Step-by-step explanation:

Let the Pn and Pl stand for the number of packs of pens (Pn) and pencils (Pl) purchased.

We know that each pack of pens is $3 and each pack of pencils is $2.50.

We are also told that Pn + Pl = 13 [In total, 13 packs were purchased].

And we're told that $35 was spent.

The $35 must equal: $3*Pn + $2.5*Pl [the sum of the number of packs of each times the price per pack for each]

3*Pn + 2.5*Pl = 35

Rearrange the first equation:

Pn + Pl = 13

Pn = 13 - Pl

Use this definition of Pn in the second equation:

3*Pn + 2.5*Pl = 35

3*(13 - Pl) + 2.5*Pl = 35

39 - 3Pl + 2.5Pl = 35

-0.5Pl = -4

Pl = 8

8 packs of pencils (Pl) were purchased.

This means the packs of pens purchased must be 5, since a total of 13 packs were purchased.

Check:

Does 5 packs of pens and 8 packs of pencils total $35?

5*($3) + 8*($2.5) = $35 ??

$15 + $20 = $35 ?

YES

You bought 5 packs of pens and 8 packs of pencils.

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Graph the exponential function.

lara31 [8.8K]

Exponential Function doesn't have a graph that intercepts x-axis only if it is f(x) = a^x (when a>0 and a ≠ 1)

If you want to plot five points to graph the function then. I'd rather do from negative number to positive number.

Substitute from -3≤x≤1 (We get 5 points)

f(-3) = 4^-3

f(-3) = 1/4^3

f(-3) = 1/64 (First point) (-3,1/64)

f(-2) = 4^-2

f(-2) = 1/4^2 = 1/16 (Second point) (-2,1/16)

f(-1)=4^-1 = 1/4 (Third point) (-1, 1/4)

f(0) = 4^0 = 1 (Fourth point) (0, 1)

f(1) = 4^1 = 4 (Fifth point) (1,4)

4 0

3 years ago

Last Wednesday, students could choose ham or turkey sandwiches for lunch. The cafeteria made 120 sandwiches in all, 30% of which

m_a_m_a [10]

Answer:

B, 36 sandwiches

Step-by-step explanation:

120/100 is equal to 1.2. so 1% is 1.2. then multiply that by 30, and you get 36 sandwiches

6 0

3 years ago

Read 2 more answers

n is a positive integer with exactly two different divisors greater than 1, how many positive factors does n^2 have? A. 4 B. 5 C

Evgesh-ka [11]

Answer:

C. 5

Step-by-step explanation:

Remember that for any integer $n$ , the integers $1 \text{and} n$ are both divisors (or factors) of $n$ . First, we will prove that n is a square, and then we will compute the factors of the n².

In this case, the integer n has exactly two different divisors greater than 1. It's impossible that $n=1$ , since 1 doesn't have positive factors greater than 1. Then $n>1$ , therefore $n$ itself is one of the required divisors. Denote by $a$ the other divisor greater than 1, and note that to satisfy the condition on the divisors, $a$ .

Because a divides n, there exists some integer $k$ such that $n=ak$ . We must have that $k>1$ , if not, then $k\leq 1$ , which implies that $ak=n\leq a$ , which contradicts the part above.

Now, $k>1$ and, by definition of divisibility, k divides n. Then k must be equal either to n or a, since we can't have three different divisors of this kind. If $k=n$ then $n=an$ and by cancellation, $1=a$ which is a contradiction. Therefore $k=a$ and $n=a^2$ .

We have that $n=(a^2)^2=a^4$ . We can write n as $n=a^3a=a^2a^2=a^4 \cdot 1$ . From the first equation, a divides n and a³ divides n. From the second equation, a² divides, and from the last one, 1 divides n and a⁴=n divides n. Thus n has exactly 5 positive factors.