MATHEMATICS HIGH SCHOOL

Find the LCM of the set of polynomials.

4m^3p, 9mp^4, 18m^4p^2

Answers

Answer 1
Answer: The LCM would be m^3p
Answer 2
Answer:

Answer:

mp3 your welcome

Step-by-step explanation:

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(45 POINTS) Answer the Question below:
4. Which cannot describe a system of linear equations?A) no solutionB) exactly two solutionC) infinite solutionsD) exactly one solution
How much farther is earth form the sun than it is from Venus?
Cos theta = -5/6180° <0<270°
Write an expression to represent: The sum of one and the product of one and a number x

A circle has a circumference of 12.It has an arc length 11. What is the central angle of the arc,in degrees?

Answers

Answer:330^(\circ)

Step-by-step explanation:

Given

Circumference of circle C=12\ units

Arc length l=11\ units

we know arc length is given by

l=(\theta )/(360^(\circ))* 2\pi r

substituting value we get

11=(\theta )/(360)* 12

(11)/(12)=(\theta )/(360)

\therefore \ \theta =330^(\circ)

Factor the trinomial
12x^2+25x+2=
factor completely
help plz

Answers

12x^2+25x+2 \n \n 12x^2 + 24x + x + 2 \n \n 12x(x + 2) + (x + 2) \n \n (x + 2)(12x + 1) \n \n

The final result is: (x + 2)(12x + 1).

12x^2+25x+2

Factor

( 12x+1)(x+2)


I hope that's help ! Good night !

What is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C?. . (Given: R = 0.08205 l · atm/mol · K). . 2.81 liters. . 4.39 liters. . 6.19 liters. . 9.67 liters

Answers

T = 227 ° C + 273 = 500 K
V = R n T/p = (0.08205 l atm/mol K * 0.8 mol* 500 K) / 5.3 atm = 6.19 lit
Answer: C ) 6.19 liters

Answer:

6.19 L

Step-by-step explanation:

We are given that

Number of moles of argon gas=0.8 moles

Pressure =P=5.3 atm

Temperature =T=227 degree Celsius=227+273=500k

We have to find the volume of the container needed to store the argon gas.

We know that

PV=nRT

Where R=0.082051 L atm/ mol k

Substitute the values in the given formula

5.3* V=0.8* 500* 0.82051

V=(0.8* 0.082051* 500)/(5.3)=6.19 L

Hence, the volume of container needed to store the gas=6.19 L

So, I really need help with functional notation...If anyone could help me solve these problems, I'd be extremely grateful!!Given the function G ( x ) =sqrt 2x-1 , find the value G (1), G(0), G(5), G(-4).
Given the function F ( x ) = x^2 - 3 x + 1, find the value F (-3), F(0), F(1).
Given the function H ( x ) = 3 x - 1, find the value H (2).

Answers

g(x)=√(2x-1)
g(1)=√(2(1)-1)=√(2-1)=√(1)=1
g(0)=√(2(0)-1)=√(0-1)=√(-1)=i
g(5)=√(2(5)-1)=√(10-1)=√(9)=3
g(-4)=√(2(-4)-1)=√(-8-1)=√(-7)=i√(7)

f(x)=x^(2)-3x+1
f(-3)=-3^(2)-3(-3)+1=9-(-9)+1=19
f(0)=0^(2)-3(0)+1=1
f(1)=1^(2)-3(1)+1=-1

h(x)=3x-1
h(2)=3(2)-1=5

Can somebody please help me with this question. Giving lots of points.

Answers

Answer:

x-multiple choice

y- short response

23x+10y=86

28x+5y=76 /×(-2)

23x+10y=86

-56x-10y=-152

-33x=-66

x=2

10y=86-46

10y=40

y=4

multiple choice=2 points

short responce =4points

Cosx-Cos2x=0
Solve. I'm not sure if I'm getting the right answers

Answers

cos(2x)= 2cos^2(x)-1\n \n cos(x)-cos(2x)=0 \n cos(x)-(2cos^2(x)-1)=0 \n -2cos^2(x)+cos(x)+1=0 \n 2cos^2(x)-cos(x)-1=0 \n (2cosx+1)(cosx-1)=0 \n \n 2cos(x)+1=0 \n cos(x)= - (1)/(2) \n x= \pm arccos(- (1)/(2) )+2 \pi k~~~k\in Z \n arccos(- (1)/(2) )= (2 \pi )/(3) \n x=\pm (2 \pi )/(3)+2 \pi k~~~k\in Z \n \n


cos(x)-1=0 \n cos(x)=1 \n x= 2\pi k~~~k \in Z
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