Answer:
Step-by-step explanation:
#1:

division sign means that we flip the fraction

now we can multiply all the constants together and all variables

now we can combine all the parts

#2:




Answer:

Step-by-step explanation:

Expand parentheses:

Subtract 42 from both sides:

Divide both sides by 14:

Hope this helps!
The angle required will be given by:
x=52/5π-5*2π=2/5π
the positive angle less than one revolution around the unit circle that os co-terminal with angle of 52/5π is 2/5π
Answer:
Answer:
\{ {{20x+30y=280} \atop {y=4x}} .{
y=4x
20x+30y=280
.
Where xx is the number of small boxes sent and yy is the number of large boxes sent.
Step-by-step explanation:
Let be xx the number of small boxes sent and yy the number of large boxes sent.
Since each small box can hold 20 books (20x20x ), each large box can hold 30 books (30y30y )and altogether can hold a total of 280 books, we can write the following equation to represent this:
20x+30y=28020x+30y=280
According to the information provided in the exercise, there were 4 times as many large boxes sent as small boxes. This can be represented with this equation:
y=4xy=4x
Therefore, the system of equation that be used to determine the number of small boxes sent and the number of large boxes sent, is:
\{ {{20x+30y=280} \atop {y=4x}} .{
y=4x
20x+30y=280
.