Answer:
<u> $50</u> is <u>27.02%</u> of <u>$185</u>.
Step-by-step explanation:
Money spent on clothing = $50
Money spent on Groceries = $100
Money spent on Gas = $35
Total money spent = $185
Our question is to find the percentage of total money that is spent on clothing.
Percentage money on clothing =
× 100
=
× 100
= 27.02% (approx)
Hence,<u> $50</u> is <u>27.02%</u> of <u>$185</u>.
Using an linear function, we find that by 2020 only 11% of all American adults believe that most qualified students get to attend college.
-----------------------------------------
A decaying linear function has the following format:

In which
- A(0) is the initial amount.
- m is the slope, that is, the yearly decay.
- In 2000, 45% believed, thus,

- Decaying by 1.7 each year, thus
.
The equation is:

It will be 11% in t years after 2000, considering t for which A(t) = 11, that is:




2000 + 20 = 2020
By 2020 only 11% of all American adults believe that most qualified students get to attend college.
A similar problem is given at brainly.com/question/24282972
5 miles high is one of the sides of a triangle depending on accuracy level
h^2=x^2+y^2
we don't have 2 distances
Tan A=O/a
O=a tan A
We solve for O because the angle is at the top of the line going up and we want the opposite angle that is along the ground
O=5×tan(173.7/2)=90.854033512
The distance he can see is:
90.85*2~181.7 miles
Now we need to find the distance between lines:
The north south distance between each line is 69 miles
thus the number of degrees he will see will be:
181.7/69
=2 19/30
Answer:
39 touchdowns and 13 fieldgoals
Step-by-step explanation:
Let t= touchdowns
f = fieldgoals
They scored 35 times
t+f = 35
Touchdown is 7 pts and fieldgoal is 3 pt
7t+3f = 193
Multiply the first equation by-7
-7t -7f =-245
Add this to the second equation
-7t -7f =-245
7t+3f = 193
----------------------
-4f =-52
Divide by -4
-4f/-4 = -52/-4
f = 13
Now we can find t
t+f = 35
t+13 = 52
Subtract 13 from each side
t+13-13 =52-13
t =39
An=a1+d(n-1)
a1=first term
d=common difference
n=which term
common difference is 5 first term is 2
an=5+2(n-1)
19th term is n=19
a19=5+2(19-1)
a19=5+2(18)
a19=5+36
a19=41
19th term is 41