Need answers for #3 please hep

R^2+3r+5-r^3/r is a quadratic/cubic/linear polynomial?

Sergeu [11.5K]

$r^2+3r+5-r^3/r$ is a linear polynomial

S<u>tep-by-step explanatio</u>n:

based on the highest power of the variables of a polynomial, it can be classified as linear,polynomial, quadratic polynomial,cubic polynomial etc.

In a linear polynomial, the highest power of the variable is 1. The general form of a linear polynomial is ax+b

In a quadratic polynomial, the highest power of the variables is 2.The general form of quadratic polynomial is $ax^2+bx+c$

In a cubic polynomial,the highest power of variable is 3.The general form of cubic polynomial is $ax^3+bx^2+cx+d$ .

Take the given question and simplify it

$r^2+3r+5-r^3/r=r^2+3r+5-r^2\\=3r+5$

It is of the form ax+b. Hence it is a linear polynomial.

3 0

4 years ago

Find the area of a trapezoid with bases 14 cm and 18 cm and height 10 cm.

4vir4ik [10]

You should add the two bases and muliply them with the height and then divide with 2
(14+18)×10÷2=160

8 0

4 years ago

50pts.

Cerrena [4.2K]

Answer:

its the 3rd 1

Step-by-step explanation:

8 0

4 years ago

How many triangles can be constructed with sides measuring 5 m, 16 m, and 5 m?

lidiya [134]

Q. How many triangles can be constructed with sides measuring 5 m, 16 m, and 5 m?

Solution:

Here we are given with the sides of the triangle as 5m, 16m and 5.

As the Triangle inequality we know that

The sum of the length of the two sides should be greater than the length of the third side. But this inequality fails here.

Hence no triangle can be made.

So the correct option is None.

Q.How many triangles can be constructed with sides measuring 6 cm, 2 cm, and 7 cm?

Solution:

Here we are given with the sides of the triangle as 6m, 2m and 7m.

As the Triangle inequality we know that

The sum of the length of the two sides should be greater than the length of the third side. The given values follows the triangle inequality.

Hence one triangle can be formed.

So the correct option is one.

8 0

3 years ago

In the adjoining figure, XY = XZ . YQ and ZP are the bisectors of <img src="https://tex.z-dn.net/?f=%20%5Cangle" id="TexFormula1

svetlana [45]

Answer:

See Below.

Step-by-step explanation:

Statements: Reasons:

$1)\, XY=XZ$ Given

$2) \text{ $ m\angle Y= m\angle Z$}$ Isosceles Triangle Theorem

$\displaystyle 3) \text{ $m\angle Y=m\angle XYQ + \angle QYZ$}$ Angle Addition

$\displaystyle 4)\text{ $YQ$ bisects $\angle XYZ$}$ Given

$5) \text{ $m\angle XYQ=m\angle QYZ$}$ Definition of Bisector

$\displaystyle 6)\text{ $m\angle Y=2m\angle QYZ$}$ Substitution

$7)\text{ $m\angle Z=m\angle XZP+m\angle PZY$}$ Angle Addition

$8)\text{ $ZP$ bisects $\angle XZY$}$ Given

$\displaystyle 9) \text{ $m\angle XZP=m\angle PZY$ }$ Definition of Bisector

$\displaystyle 10) \text{ $ m\angle Z = 2m\angle PZY $}$ Substitution

$11)\text{ } 2m\angle QYZ=2m\angle PZY$ Substitution

$12)\text{ }m\angle QYZ=m\angle PZY$ Division Property of Equality

$13)\text{ } YZ=YZ$ Reflexive Property

$14)\text{ } \Delta YZP\cong\Delta ZYQ$ Angle-Side-Angle Congruence*

$15)\text{ } YQ=ZP$ CPCTC

*For clarification:

∠Y = ∠Z

YZ = YZ (or ZY)

∠PZY = ∠QYZ

So, Angle-Side-Angle Congruence:

ΔYZP is congruent to ΔZYQ

5 0

3 years ago

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