The function's constant term is 5. Step-by-step explanation: we have where b represents the y-intercept or the constant term of the function. The x-intercept indicates the value of x when the function equals zero. Thus, for x = -3, f(x) = 0 and for x = -5, f(x) = 0. Substitute either of the intercepts into the function. Check the other intercept as well. For x = -5, this is proven to be true. Hence, the consistent term in the function is 5.
According to the rule of 72,
72/rate=time
72÷9.6=7.5 years
An alternative method for resolution using the main formula
2300=1150(1+0.096/4)^4t
Isolate t
t=((log(2,300÷1,150)÷log(1+(0.096÷4))÷4))=7.31 years
I hope this is helpful:-)
Answer:
B).35
Step-by-step explanation:
The Central Limit Theorem asserts that for a normally distributed variable X, with mean 
and standard deviation 
, sample means' distribution with a size n can be closely approximated by a normal distribution with mean and standard deviation 
.
This theorem also applies to skewed variables, provided n is at least 30.
In this scenario:
Then
Therefore, the right answer is:
B).35
Response:
∠PQL=∠TRN [Angles corresponding]
Thus, PQ║RS and PQ=RS
Detailed explanation:
The side PQ has been drawn.
A second side QR is traced, forming an acute angle with side PQ.
Now side QR is extended to the left.
Create an arc from point Q such that it intersects QP at M and extends RQ at L. Without altering the compass width (i.e., the distance between the nib and pencil), draw an arc from R to intersect RQ at N. Now measure the distance LM with a compass. Position the compass at N and mark an arc cut from point R. Designate this intersection as T. Draw a line from point R through T. Then measure the length of PQ with the compass. Position your compass at R and create an arc on the produced line RT at S. Thus, we ascertain that PQ║RS and PQ=RS.
This occurs because
∠MQL=∠NRT [corresponding angles, with QR acting as the transversal]
∵PQ║RS and PQ=RS [This identifies PQRS as a parallelogram]
Out of the four students who illustrated their explanations
Student 2 presented a partially correct but valid explanation.
