305 to the nearest ten is 310.
Answer:
2. Judy =$5
Ben= $4
Step-by-step explanation:
2. Let Judy = x and Ben = y
8x + 10y = 80
9x + 5y = 65
Solve these simultaneous equations.
8
x + 10
y = 80
18
x + 10
y = 130
Take the second equation away from the first equation
−
10
x = −
50
x = 5
This means that Judy gets paid $5 an hour.
Therefore, Ben gets paid $4 an hour.
Answer:
3. 24 quarters and 16 dimes
Step-by-step explanation:
3. Let the number of dimes = x and the number of quarters = y
Value Value
Type Number of of
of of EACH ALL
coin coins coin coins
-------------------------------------------
dimes x $0.10 $0.10x
quarters y $0.25 $0.25y
-------------------------------------------
TOTALS 40 $7.60
x + y = 40
0.10x + 0.25y = 7.60
Get rid of decimals by multiplying every term by 100:
10x + 25y = 760
So we have the system of equations:
x + y = 40
10x + 25y = 760
We solve by substitution. Solve the first equation for y:
x + y = 40
y = 40 - x
Substitute (40 - x) for y in 10x + 25y = 760
10x + 25(40 - x) = 760
10x + 1000 - 25x = 760
-15x + 1000 = 760
-15x = -240
x = 16 = the number of dimes.
Substitute in y = 40 - x
y = 40 - (16)
y = 24 quarters.
$1.60 + $6.00 = $7.60
I hope that helped, sorry for taking so long :-)
Answer:
The height of the lighthouse is 
Step-by-step explanation:
Let
h -----> the height of the lighthouse
we know that
The tangent of angle of 68 degrees is equal to divide the height of the lighthouse by the horizontal distance from the buoy to the base of lighthouse
so
Solve for h
What triangle? Is there one on here?
Answer:
This problem is incomplete, we do not know the fraction of the students that have a dog and also have a cat. Suppose we write the problem as:
"In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. X of the students who have a dog as a pet also have cat as a pet. If there are 45 students in her class, how many have both a dog and a cat as pets?"
Where X must be a positive number smaller than one, now we can solve it:
we know that in the class we have 45 students, and 4/5 of those students have dogs, so the number of students that have a dog as a pet is:
N = 45*(4/5) = 36
And we know that X of those 36 students also have a cat, so the number of students that have a dog and a cat is:
M = 36*X
now, we do not have, suppose that the value of X is 1/2 ("1/2 of the students who have a dog also have a cat")
M = 36*(1/2) = 18
So you can replace the value of X in the equation and find the number of students that have a dog and a cat as pets.
