Answer:
2 erasers cost <u><em>$1.20</em></u>
Step-by-step explanation:
Nigel bought 10 pencils and 5 erasers for $8.
We can write this algebraically:
10p + 5e = 8
We know that the price of 2 pencils is $1. That is $0.50/Pencil.
So, p = $0.50. Now we can plug that into out equation and find the price of one eraser:
10(0.50) + 5e = 8
5 + 5e = 8
5e = 3
e = $0.60
Now, we have the price of one eraser,<em> </em><u><em>$0.60</em></u>, and we can multiply this by 2 to find the price of 2 erasers, which equates to <u><em>$1.20</em></u>.
So use the <span>Pythagorean Theorem. A^2 + B^2 = C^2 10^2 +12^2 = 15.6 hope this helps.
</span>
Answer:
1/8
Step-by-step explanation:
Let
The total fraction of the book read = 1
The remaining 4 days be represented as= x
Hence:
1 = Fraction of book read on Monday + Fraction of the book read on Tuesday + Fraction of the book on the remaining 4 days
1 = 1/6+ 1/3 + x
x = 1 - (1/6+ 1/3)
x = 1 - (1 + 2/6)
x = 1 - (3/6)
x = 1 - 1/2
x = 1/2
The fraction of the book read in 4 days = 1/2
To calculate the fraction of the book read on each of the days, since he read the same number of pages on each of the days we have:
1/2 ÷ 4 days
= 1/2 × 1/4
= 1/8
Therefore, the fraction of the book that was read on each of the 4 days is 1/8
<h3>
Answer:</h3>
A: see below
B: no
<h3>
Step-by-step explanation:</h3>
Part A. The first equation graphs as the area below the dashed line with y-intercept -7 and slope 2.
The second equation graphs as the area above the solid line with x-intercept 6 and y-intercept 3.
The doubly-shaded area representing the solution space is that space approximately the upper-right quadrant of the four sections of the coordinate plane created by the intersection of the lines.
Part B. The point (3, -7) is in the lower-right quadrant of the sections of the coordinate plane described in part A. Thus it is NOT A SOLUTION.
The point (3, -7) fails to satisfy the second inequality. That is ...
-7 ≥ -1/2·3 +3 = 3/2 . . . . is NOT TRUE
In order to be part of the solution space, a point must satisfy <em>both</em> inequalities.