Answer:
$9+j×0.5
Step-by-step explanation:
0.5$ each cup
Cup sold j
0.5×j= $9+j×0.5 or $9+j.05
Answer:
No because 50 cm would be short of 14 cm required to cover the total perimeter
Step-by-step explanation:
2 meter long string into 4 equal part
1 part = 50 cm
Side length of the square = 16
16 × 4 = 64cm ⇔ Perimeter of the square box
50 < 64
Let's let x equal the original price of the dress.
We know that 16.6666% is equal to 0.16666 in decimal form by dividing the percent by 100.
So we can set up the equation:
because we know the discount, and we know that the discount of the original price was subtracted from the original price.
So then let's simplify by taking out x, the common factor:
Then solve for x:
So now we know that the original price of the dress was $39.
Answer:
98 bracelets
Step-by-step explanation:
Dorinda makes 50 bracelets (5*8+10=50) and Evan makes 48 (6*8=48). 48+50=98
Answer:
- 13 - 23 = - 13 + (-23)
= -36
- 13 - (- 23) = - 13 + 23
= 10
Step-by-step explanation:
In any order of operation (i.e. addition, subtraction, multiplication, devision) the following rules apply for two values for instance <em>a </em>and <em>b </em>where <em>a > b </em>:
→ <em>If +a and +b then the final sign will be + (positive)</em>
→ <em>If +a and -b then the final sign will be + (positive) since a>b</em>
→ <em>If -a and +b then the final sign will be - (negative) since a>b</em>
→ <em>If -a and -b then the final sign will be - (negative)</em>
In the given question we have two cases.
Case 1:
- 13 - 23 = - 13 + (-23) <em>→ since it is given that we add a negative value of -23, therefore it keeps its sign (so this would be a negative addition i.e. adding two negative values)</em>
Thus we have:
- 13 - 23 = - 13 + (-23)
= -36
Case 2:
- 13 - (- 23) = - 13 + 23 <em>→ since we stated earlier that - and - gives positive sign (i.e. we subtract two negative values)</em>
Thus we have:
- 13 - (- 23) = - 13 + 23
= 10
<em>So we can see that: </em>
<em>Adding two Negative values gives a Negative value (Case 1)</em>
Subtracting two Negative values gives a Positive value (Case 2)
