Answer:
3) girls = 24% = 456
let the total number of students be = x
the total number of students is 1900
4) increase = 20%
new salary = 24000 rupees
let his original salary be = y
his original salary is 20,000 rupees
At time t= 0, Elevator A is 40 feet above the ground floor.
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The amount of oxygen in the flask through the intake tube illustrates proportions
The flask 99.35% will contain of oxygen after 5L have passed through the intake tube
<h3>How to calculate the percentage of oxygen?</h3>
To do this, we make use of the following representations
- P for the proportion of oxygen
- V for te volume of oxygen
- x for the proportion of oxygen in the flask after 5 L passed through
So. we have the following derivative
dP = dV - P * dV
Divide through by dV
dP/dV = 1 - P
Next, we set up the integral
Integrate both sides
Expand
-ln(|x - 1|) + ln(|0.04 - 1|) = 5
Evaluate the difference
-ln(|x - 1|) + ln(|-0.96|) = 5
Express as a fraction
ln(0.96/|x - 1|) = 5
Express both sides as exponents with a base of 2
e^ln(0.96/|x - 1|) = e^5
Solve for x
x = 1 - 0.96/e^5
Evaluate the difference
x = 0.9935
Express as percentage
x = 99.35%
Hence, the flask 99.35% will contain of oxygen after 5L have passed through the intake tube
Read more about proportions at:
First, "graph" means graph of the given function f(x). It is not necessary to include the graph with the question because at this level, either the user can "see" the graph mentally, or use technology (calculator or software) to graph the function.
(a) The Leading coefficient test for polynomial functions
The test is as follows. The sign of the leading coefficient (LC, coefficient of the highest power of the variable) together with the degree (odd or even) of the polynomial function tells us the end behaviour of the function according to the following table:
LC/degree Even Odd
Positive +inf/+inf -inf/+inf
Negative -inf/-inf +inf/-inf
It can be easily memorized by comparing with a line or a parabola
y=x (positive leading coefficient, odd degree) goes from -inf to +inf
y=x^2 (positive leading coefficient, even degree) goes from +inf to +inf.
Any negative leading coefficient will just invert the signs.
For the given polynomial function
f(x) = x^2(x+2) = x^3+x^2
The leading coefficient is +1 (positive), and degree is three (odd)
So it behaves the same way as a straight line y=x, so the end behviours are
{-inf, +inf} as x-> {-inf, +inf} respectively.
(b) x-intercepts
Since f(x)=x^2(x+2), the x-intercepts are x=0 (multiplicity 2) and x=-2.
=>
From the left, graph is increasing and crosses x=-2 once from negative to positive, after which graph is decreasing. At x=0, because of the root with multiplicity 2, graph touches the origin (0,0) and turns back towards +inf.
(c) the y-intercept
from the polynomial f(x)=x^2(x+2)=x^3+x^2 = x^3+x^2+0,
we see that the y-intercept =0 because the "initial value" or the constant term is zero.
(a) The Leading coefficient test for polynomial functions
The test is as follows. The sign of the leading coefficient (LC, coefficient of the highest power of the variable) together with the degree (odd or even) of the polynomial function tells us the end behaviour of the function according to the following table:
LC/degree Even Odd
Positive +inf/+inf -inf/+inf
Negative -inf/-inf +inf/-inf
It can be easily memorized by comparing with a line or a parabola
y=x (positive leading coefficient, odd degree) goes from -inf to +inf
y=x^2 (positive leading coefficient, even degree) goes from +inf to +inf.
Any negative leading coefficient will just invert the signs.
For the given polynomial function
f(x) = x^2(x+2) = x^3+x^2
The leading coefficient is +1 (positive), and degree is three (odd)
So it behaves the same way as a straight line y=x, so the end behviours are
{-inf, +inf} as x-> {-inf, +inf} respectively.
(b) x-intercepts
Since f(x)=x^2(x+2), the x-intercepts are x=0 (multiplicity 2) and x=-2.
=>
From the left, graph is increasing and crosses x=-2 once from negative to positive, after which graph is decreasing. At x=0, because of the root with multiplicity 2, graph touches the origin (0,0) and turns back towards +inf.
(c) the y-intercept
from the polynomial f(x)=x^2(x+2)=x^3+x^2 = x^3+x^2+0,
we see that the y-intercept =0 because the "initial value" or the constant term is zero.