Answer:
Step-by-step explanation:
we are given half-life of PO-210 and the initial mass
we want to figure out the remaining mass <u>after</u><u> </u><u>4</u><u>2</u><u>0</u><u> </u><u>days</u><u> </u>
in order to solve so we can consider the half-life formula given by
where:
- f(t) is the remaining quantity of a substance after time t has elapsed.
- a is the initial quantity of this substance.
- T is the half-life
since it halves every 140 days our T is 140 and t is 420. as the initial mass of the sample is 5 our a is 5
thus substitute:
reduce fraction:
By using calculator we acquire:
hence, the remaining sample after 420 days is 0.625 kg
For number 25 it’s x=4 and for 26 it’s x=7
5 is less than or equal to -3+x
The value of x is 5/8.
<u>Step-by-step explanation</u>:
Given,
- The lines PQ and RS are parallel to each other.
- slope of PQ= x-1/4
- slope of RS = 3/8
The slopes of parallel lines are equal.
slope of PQ = slope of RS
⇒ x-1/4 = 3/8
⇒ (4x-1)/4 = 3/8
⇒ 8(4x-1) = 4(3)
⇒ 32x-8 = 12
⇒ 32x = 20
x = 20/32
x = 5/8
Answer:
Step-by-step explanation:
a²+b² = c² c is always the longest line thats opposite to the 90 degree angle
for 1
15² + b² = 17²
b² = 17² - 15²
b = √(17² - 15²)
b= 8