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2 years ago

Find the degree of the monomial. -20.5

Mathematics

2 answers:

Nadya [2.5K]2 years ago

7 0

Answer:

Step-by-step explanation:

-20.5 is the same as -20.5 $x^{0}$ , since anything to the zero power is equal to 1.

Nookie1986 [14]2 years ago

4 0

Answer:

The degree of a monomial is the sum of the exponents of the variables.

Therefore,

The degree of the monomial - 20.5 is 0, because there is no variables only numeric terms.

Step-by-step explanation:

To find the degree of monomial.

Basically,

- If one term is present, then it is monomial.

- If two term is present, then it is binomial.

- If three term is present, then it is trinomial.

- If more than three term is present, then it is polynomial.

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3 years ago

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3 years ago

State whether the following coordinates on a Cartesian plane form an acute, obtuse or right triangle: a) (-1, 1), (7,-2) and (1,

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Answer:

a) Acute triangle

b) Right triangle

Step-by-step explanation:

∵ A triangle having sides a, b and c is called,

Acute : If a² + b² > c² or a² + c² > b² or b² + c² > a²,

Obtuse : if a² + b² < c² or a² + c² < b² or b² + c² < a²

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a) Let A≡(-1, 1), B≡(7,-2) and C≡(1,-5),

By the distance formula,

$AB=\sqrt{(7-(-1))^2+(-2-1)^2}=\sqrt{(7+1)^2+(-3)^2}=\sqrt{8^2+3^2}=\sqrt{64+9}=\sqrt{73}\text{ unit}$

Similarly,

$BC=\sqrt{45}\text{ unit}$

$CA=\sqrt{40}\text{ unit}$

∵ The sum of any two sides is greater than third side,

So, ABC is a triangle,

Now,

$AB^2 + BC^2 > CA^2$

⇒ ABC is an acute triangle.

b) Let P≡(0,6), Q≡(1,2) and R≡(5,3),

By the distance formula,

PQ = √17 unit,

QR = √17 unit,

RP = √34 unit,

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⇒ PQR is a triangle ,

Also,

$RP^2=PQ^2+QR^2$

Hence, PQR is a right triangle.

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3 years ago

Find the degree of the monomial. -20.5​

Find the degree of the monomial. -20.5