Answer:
0.05n+0.10d=1.90
n=d+2
Step-by-step explanation:
One nickel is worth $0.05, so n nickels are worth 0.05n. One dime is worth $0.10, so dd dimes are worth 0.10d.The total 0.05n+0.10d equals $1.90:
0.05n+0.10d=1.90
Since there are 2 more nickels than dimes, there are more nickels, so if we add 2 to the number of dimes, we will get the number of nickels, meaning nn equals d+2.d+2.
n=d+2
From the problem, we create a system of equations that represent the value and quantity of nickels and dimes Victoria has. The system is: 0.05N + 0.10D = 1.90 and N = D + 2, where N represents nickels and D represents dimes.
The problem is about finding the number of nickels and dimes Victoria has. To do this, we will use a system of equations. We'll define N as the number of nickels and D as the number of dimes.
The first equation is based on the total value: since a nickel is worth $0.05 and a dime is worth $0.10, Victoria's coins add up to $1.90 - hence 0.05N + 0.10D = 1.90
The second equation is based on the quantity of coins: Victoria has 2 more nickels than dimes - hence N = D + 2.
Therefore, the system of equations is:
#SPJ11
Answer:
132 cm squared
Step-by-step explanation:
Find the area of the rectangle on the top first.
Finding the area of a rectangle:
The formula to find the area of a rectangle is to multiply the length by the width. So for your rectangle you would do
7 x 4 = 28cm squared. Keep this is mind and we will move on the the next problem.
The figure under the rectangle is a trapezoid.
To find the area of a trapezoid:
Step 1)
Average the bases.
The dotted line is 7 cm. This is a base.
The line on the very bottom is 19 cm. This is your 2nd base. Find the average of these bases.
(7 + 19) / 2
= 26/2
=13
Step 2)
Multiply this by the height. The height of the trapezoid is 8 cm.
Multiply:
13 x 8
This would get you 104. This is the area of the trapezoid.
Add 104 to the area of the rectangle, 28, to get 132.
The area of the whole figure, therefore, is 132cm squared.
I hope this helps!
It is -25 or just 25
5/s+2/5=2/s
Answer:
Step-by-step explanation:
Errors in Algebraic Operations
It's usual that students make mistakes when misunderstanding the application of algebra's basic rules. Here we have two of them
The first expression is
Let's arrange into format:
We can clearly see in all of the factors in the expression the signs were changed correctly, but the result should have been preceeded with a negative sign, because it makes 3 (odd number) negatives, resulting in a negative expression. The correct form is
Now for the second expression
Let's arrange into format
It's a clear mistake because it was asssumed a product of fractions instead of a SUM of fractions. If the result was correct, then the expression should have been