Answer:
147.5 km and 64.4 km
Step-by-step explanation:
a=120 km
b=70 km
β=28 degrees (
∘)
b^2=(a^2)+(c^2)−2ac*cosβ
70^2
=(120^2
)+(c^2)−2⋅ 120⋅ c⋅ cos(28∘ )
(c^2
) −211.907c+9500=0
note p, q, and r are replacement variables in the Pythagorean theorem since a, b, and c are already in use
p=1;q=−211.907;r=9500
D=(q^2
) −4pr=(211.907^2
)−4⋅1⋅9500=6904.75561996
D>0
=
(−q± 
)/2p=(211.91±
)/2
=105.95371114±41.5474295834
(
−147.501140726)(
−64.4062815596)=0
=147.501140726
=64.4062815596
X= -8 that uis your answer for that question :)
Let N be the number of items sold and p the price.
Since the variation is inverse, then the relation between N and p is:
For N=20000 and p = $9.5, we get the formula:
If p = 8.75, then the number of items sold can be computed using the formula:
Answer:
c) 6
Step-by-step explanation:
This is a straight-forward application of the Law of Cosines:
... a² = b² + c² -2bc·cos(A)
... a² = 8² +11² -2·8·11·cos(32.2°) ≈ 64 +121 -176·0.8462
... a² ≈ 36.07000
... a ≈ 6.00583
The best choice is c) 6.
Answer:
After 7.40 years it will be worth less than 21500
Step-by-step explanation:
This problem is solved using a compound interest function.
This function has the following formula:
Where:
P is the initial price = $ 34,000
n is the depreciation rate = 0.06
t is the elapsed time
The equation that models this situation is:
Now we want to know after how many years the car is worth less than $ 21500.
Then we do y = $ 21,500. and we clear t.
After 7.40 years it will be worth less than 21500
