Pls help me out with this question
2 answers:
Answer:
83 m (nearest metre)
Step-by-step explanation:
This can be modeled as a <u>right triangle</u>, where the base is 60 m and the height is the height of the building.
<u>Alternate Interior Angles Theorem (Z-angles)</u>
If a line intersects a set of parallel lines in the same plane at two distinct points, the alternate interior angles that are formed are congruent.
Using the Alternate Interior Angles theorem, the angle between the ground and the dashed line (angle of elevation) is 54°.
<u>Tan trigonometric ratio</u>
where:
is the angle
- O is the side opposite the angle
- A is the side adjacent the angle
Given:
- O = height of building (h)
- A = 60
Substituting the given values into the formula and solving for h:
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Vertical asymptotes are when y tends to infinity, which in other words means you need a fraction where you're dividing by zero. So to find the vertical asymptotes, solve the denominator of the fraction and the solutions are the asymptotes since that is where it will be dividing by zero.
Horizontal asymptotes are the opposite and is where x tends to infinity -- This can be done by dividing through with the xs still there, since you are effectively just looking at multiples of infinity and infinity squared and so on. When you have an x to the power you can forget about all the other x's on the same half of the fraction, since infinity squared is infinitely bigger than infinity - yes it's abstract and a bit stupid I know.
If the degree of the polynomial on the denominator is larger than the numerator, there is an asymptote at y=0 (since the bottom becomes infinitely larger than the top). If it is lower than the numerator, you have to divide through by long division and you end up with an oblique asymptote i.e. an asymptote where the dotted line would follow a linear equation e.g. y=x
For example
The vertical asymptote can be found by solving 2x^2-8=0 i.e. x=+-2 so our vertical asymptotes are at 2 and -2.
The horizontal asymptote, think of all the x's as infinity. SInce there are x^2s, we dont need to worry about the numbers or the other x's
Therefore we are left with 6x^2/2x^2, which gives us 3. Therefore our horizontal asymptote would be at y=3.
Horizontal asymptotes are the opposite and is where x tends to infinity -- This can be done by dividing through with the xs still there, since you are effectively just looking at multiples of infinity and infinity squared and so on. When you have an x to the power you can forget about all the other x's on the same half of the fraction, since infinity squared is infinitely bigger than infinity - yes it's abstract and a bit stupid I know.
If the degree of the polynomial on the denominator is larger than the numerator, there is an asymptote at y=0 (since the bottom becomes infinitely larger than the top). If it is lower than the numerator, you have to divide through by long division and you end up with an oblique asymptote i.e. an asymptote where the dotted line would follow a linear equation e.g. y=x
For example
The vertical asymptote can be found by solving 2x^2-8=0 i.e. x=+-2 so our vertical asymptotes are at 2 and -2.
The horizontal asymptote, think of all the x's as infinity. SInce there are x^2s, we dont need to worry about the numbers or the other x's
Therefore we are left with 6x^2/2x^2, which gives us 3. Therefore our horizontal asymptote would be at y=3.
Answer:
The correct answer is: Option 1: $19.50
Step-by-step explanation:
Simple interest is given by the formula:
Here
P is the principal amount
r is the interest rate
and t is the time in years
Given
P = 300
r = 6 1/2 = 6.5%
t = 1
Putting the values in the formula, we get
The simple interest paid to the nearest cent is 19.50 dollars.
Hence,
The correct answer is: Option 1: $19.50
Answer:
The equation of the quadratic in standard form is:
Step-by-step explanation:
Since they give us the information about where the vertex of the parabola is located, and one extra points where it passes through, we can use the general form of a quadratic in vertex form:
where is the location of the vertex (in our case the point (-2,6).
Therefore the equation above becomes:
Now,we can use the fact that the point (-4,-2) is also a point of the graph, to find the value of the parameter :
Then, the equation of the quadratic with such characteristics becomes:
which is the equation of the quadratic in standard form.