# Use function notation to describe the way the second variable (DV) depends upon the first variable (IV). Determine the domain and range for each, dete

Use function notation to describe the way the second variable (DV) depends upon the first variable (IV). Determine the domain and range for each, determine if there is a positive, negative, or no relationship, and explain your answers.
A)IV: an acute angle V in a right triangle: DV: the area B of the triangle if the hypotenuse is a fixed length G.
B)IV: one leg P of a right triangle: DV: the hypotenuse G of the right triangle if the other leg is 2
C)IV: the hypotenuse G of a right triangle: DV: the other leg P of the right triangle is one leg is 5.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

diskusje5
A) In this part, we are given an independent variable – an acute angle V in a right triangle, dependent variable - area B of the triangle and a constant – hypotenuse G has a fixed length.
We can first express the two legs of the right triangle in terms of hypotenuse G and acute angle V as shown below:
Height $=G\mathrm{sin}\left(V\right)$
Base $=G\mathrm{cos}\left(V\right)$
Use the formula for area of triangle as shown below:
Area $=\frac{1}{2}$ *Base*Height
Area $=\frac{1}{2}\cdot \left(G\mathrm{cos}\left(V\right)\right)\cdot \left(G\mathrm{sin}\left(V\right)\right)$
Area $=\frac{1}{2}{G}^{2}\mathrm{sin}\left(V\right)\mathrm{cos}\left(V\right)$
Area $=\frac{1}{4}{G}^{2}\mathrm{sin}\left(2V\right)$
Domain of this function is $\left(0,\frac{\pi }{2}\right)$ and range is $\left(0,\frac{{G}^{2}}{4}\right)$.
Since area function is oscillating function, therefore, the relation between two variables is neither positive and nor negative.
B) In this part, we are given an independent variable – a leg P of a right triangle, dependent variable - the hypotenuse G of the right triangle, and constant – second leg of length 2.
We can use Pythagorean theorem to express the relationship between P, G and 2 as shown below:
${P}^{2}+{2}^{2}$
${G}^{2}={P}^{2}+4$
$G=\sqrt{{P}^{2}+4}$
In reference to the given question, since P represents a leg of a right triangle, it can take any real number greater than 0. Therefore, domain of this function is $\left(0,\mathrm{\infty }\right)$.
Range of this function is all real numbers greater than 2, that is, $\left(2,\mathrm{\infty }\right)$.
Since value of hypotenuse increases as the length of leg increases, therefore, there is a positive relationship between the two variables.
C) In this part, we are given an independent variable – the hypotenuse G of a right triangle, dependent variable - the leg P of the right triangle, and constant – second leg of length 5.
We can use Pythagorean theorem to express the relationship between P, G and 2 as shown below:
${P}^{2}+{5}^{2}={G}^{2}$
${P}^{2}={G}^{2}-25$
$P=\sqrt{{G}^{2}-25}$
G can take any real number greater than 5 in order for this function to exist. Therefore, domain of this function is $\left(5,\mathrm{\infty }\right)$.
Range of this function is all positive real numbers, that is, $\left(0,\mathrm{\infty }\right)$.
Since value of leg increases as the length of hypotenuse increases, therefore, there is a positive relationship between the two variables.