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2 years ago

Miguel bought 8 equally priced pens for $3.84. How much will 15 of these pens cost?

Mathematics

1 answer:

Snezhnost [94]2 years ago

5 0

Answer:

$7.20

Step-by-step explanation:

If 8 pens cost $3.84, we need to find out how much ojne pen costs.

To find out the cost of one pen we need to divide $3.84 by 8.

$3.84 / 8 = $0.48

1 pen cost $0.48

We then need to work out the cost of 15 pens. So from using the information that one pen costs $0.48 we can multiply that by 15 to find out the answer.

$0.48 x 15 = $7.20

15 pens cost $7.20

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How can you use substitution to solve a system of linear equations?

makvit [3.9K]

Step 1:
Solve one of the equations for either x = or y = .
Step 2:
Substitute the solution from step 1 into the other equation.
Step 3:
Solve this new equation.
Step 4:
Solve for the second variable.

Example 1: Solve the following system by substitution
Substitution Method Example
Solution:
Step 1: Solve one of the equations for either x = or y = . We will solve second equation for y.
solution step 1
Step 2: Substitute the solution from step 1 into the second equation.
solution step 2
Step 3: Solve this new equation.
solution step 3
Step 4: Solve for the second variable
solution step 4
The solution is: (x, y) = (10, -5)
Hope this helps!

6 0

3 years ago

ILL GIVE BRAINLIEST!! PLS HELP!!! Ten students are taking both algebra and drafting. There are 24 students taking algebra. There

Minchanka [31]

Answer:

15 students

Step-by-step explanation:

from the algebra = 24 - 10 = 14

from the draft = 11 - 10 = 1

14 + 1 = 15

the total students that are taking algebra or drafting but not both is 15 students

4 0

3 years ago

Read 2 more answers

Let A = { x ∈ R : x 2 − 5 x + 4 ≤ 0 } , B = (3 , 5), and C = (3 , 4]. Show that A ∩ B = C . (Recall, you must show two separate

kolbaska11 [484]

Answer:

You can proceed as follows:

Step-by-step explanation:

First solve the quadratic inequality $x^{2}-5x+4\leq 0$ . To do that, factorize, then we have that $(x-4)(x-1)\leq 0$ . This implies that

$x-1\leq 0\, \text{and}\, x-4\geq 0$

$x-1 \geq 0\, \text{and}\, x-4\leq 0$

In the first case the solution is the empty set $\emptyset$ . In the second case the solution is the interval $1\leq x \leq 4$ . Now we have that

$A=[1,4]$

$B=(3,5)$

$C=(3,4]$ .

To show that $A\cap B\subseteq C$ consider $x\in A\cap B$ . Then $1\leq x \leq 4\, \text{and}\, 1$ , this implies that $3$ , then $x\in C$ . Now, to show that $C\subseteq A\cap B$ consider $x\in C$ , then $3$ , then $1\leq x \leq 4\, \text{and}\, 3$ , then $x\in [1,4] \, \text{and}\, x\in (3,5)$ , this implies that $x\in A\cap B$ .