If you assign variables to the problem, it can make things a lot simpler. Lets say chairs are x and tables are y. Therefore you have:
2x+6y=40
5x+3y=25
Now you can isolate the variable of one equation and put it into another (it doesn't matter which. I'm going to manipulate the top equation to plug into the bottom one).
2x=40-6y
x=20-3y
Now I plug into bottom equatioin:
5(20-3y) + 3y=25
100-15y+3y=25
100-12y=25
-12y=-75
y=$6.25
Now you can plug in y in either equation to get x.
2x+6(6.25)=40
37.5+2x=40
2x=2.5
x=1.25
So it costs $6.25 for each table and $1.25 for each chair. If you think about it, it would make sense for the table to cost more for the chair.
Answer:
$2327.5
Step-by-step explanation:
The stock of company A lost 2% today .
Let the price of the stock at the beginning of the day be A.
Therefore, a 2% loss of A = $46.55
That’s
2% /100% x A = $46.55
0.02 x A = $46.55
Divide both sides by 0.02
0.02/0.02 x A = $46.55/0.02
A = $2327.5
The price of stock at the beginning of the opening day was $2327.5
C. Savings account B because it has more compounding periods per year.
Step-by-step explanation:
Step 1:
Savings account A has an APR of 5% which compounds interest semiannually. This means that savings account A compounds twice in a year. If account A compounds 5% a time, it would compound 5(2) = 10% in a single year.
Step 2:
Savings account B also has an APR of 5% which compounds interest quarterly. This means that savings account B compounds four times in a year. If account B compounds 5% a time, it would compound 5(4) = 20% in a single year.
Step 3:
Savings account A gets an interest of 5% a year while savings account B gets an interest of 10% so account B offers a higher APR because of more compoundings in a year.
Answer:
The range of values for x is; 5 < x < 29
Step-by-step explanation:
The given parameters are;
= 15
= 18
= 
Given
= by reflexive property
∠BCA = 2·x - 10
∠DCA = 48°
Since 15 < 18, given the common sides of the tringle ΔABC and triangle ΔADC, angle ∡(2·x - 10)° < 48°
Therefore;
2x - 10 < 48
2·x < 48 + 10
∴ x < 29
Also, given that 2·x - 10 is real, 0 < 2·x - 10
10 < 2·x
10/2 < x
5 < x
Therefore, an acceptable range is 5 < x < 29
