Identify if each polynomial is quadratic, linear, or neither option 1) f(x)= x^2+3x+4 option 2) g(x) = 9+7x option 3) h(x)= x^-2 -36 option 4) p(x)= 4x option 5) r(x)=x^2+3 option 6)c(x)= 1500+14x

A polynomial is classified as linear or quadratic on the basis of degree of a polynomial.

The degree of a polynomial is the highest power of its monomials (individual terms) with non-zero coefficients.

If the degree of a polynomial is one(1), then the it is linear and if the degree of a polynomial is two(2), then it is quadratic.

Consider the first part.

1. f(x)=

Here, the highest power of 'x' is 2.

Therefore, its degree is 2. So, f(x) is a quadratic polynomial.

2. g(x)=

Here, the highest power of 'x' is 1.

Therefore, its degree is 1. So, g(x) is a linear polynomial.

3. h(x)=

Here, the highest power of 'x' is 2.

Therefore, its degree is 2. So, h(x) is a quadratic polynomial.

4. p(x)=4x

Here, the highest power of 'x' is 1.

Therefore, its degree is 1. So, p(x) is a linear polynomial.

5. r(x)=

Here, the highest power of 'x' is 2.

Therefore, its degree is 2. So, r(x) is a quadratic polynomial.

6. c(x)=1500+14x

Here, the highest power of 'x' is 1.

Therefore, its degree is 1. So, p(x) is a linear polynomial.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.