Answer:
The approximate area is A) 5.09cm²
Step-by-step explanation:
Area of a Circle = πr²
r = 1.8
Plug in our values
π(1.8cm²)
Evaluate the area of the entire circle
a = π(1.8cm²)
a = π(3.24cm)
a = 10.179cm²
Area of a sector, with the area of the circle of the sector being a = theta/360 * a
Evaluate the area of the sector
180/360*10.719cm²
0.5 * 10.179cm²
5.0895cm²
Round the value up
5.09cm²
Answer:
515
Step-by-step explanation:
If it's a multiple of five and not ten, that would be the units digit must be 5. Hundreds=units. 11-10=1. So the answer is 515
Answer:
6.4 m
Step-by-step explanation:
We have 2 expressions here. The first one is the fact that r = y. That's one of 2 equations. The second one involves whats' left after cutting off certain lengths of each color string. We cut 2.5 m from red, we cut 3.8 m from yellow. We know that what's left of red is 1.5 times the length of what's left of yellow. What's left of red is r - 2.5; what's left of yellow is y - 3.8. We know that r = 1.5y, so filling that in with our corresponding expressions gives us
r - 2.5 = 1.5(y - 3.8)
Distribute to get
r - 25 = 1.5y - 3.2
Now from the first expression, r = y, so fill in y for r to get an equation in one variable:
y - 2.5 = 1.5y - 3.2
Combine like terms:
-.5y = -3.2 and divide to get
y = 6.4
Check it to make sure it works. What's left of red should be 1.5 times the length of what's left of yellow and y = 6.4:
What's left of red: 6.4 - 2.5 = 3.9
What's left of yellow: 6.4 - 3.8 = 2.6
1.5 x 2.6 = 3.9, just like it should!
Ok so
t=number of text messsages
p=number of picture messages
total number is 33
t+p=33
if text is 0.18 per pic and picture is 0.45 per text
total cost is 8.3
0.18t+0.45p=8.3
times 100 both sides for ease
18t+45p=830
we now have 2 equations
t+p=33
18t+45p=830
multiply first equation by -18 and add to second equation
18t+45p=830
<span>-18t-18p=-594 +</span>
0t+27p=236
27p=236
divide both sides by 27
p=8.740740740740744
this is impossible becasue you can't send 0.740740740740744 of a message so you did a mistype somewhere
