All categories

2 years ago

What is -2m^2-2mn+8-10m+10n+mn+4n^2-5n simplify to

1 answer:

4 0

Find like terms and group them
3n and 5n are same and 4n^2 and 4n^2 are same
but 3mn and 5n are not and 3n and 2n^3 are not

group like terms
(-2m^2)+(-2mn+1mn)+(8)+(-10m)+(10n-5n)+(2n^2)
add like terms
(-2m^2)+(-mn)+(8)+(-10m)+(5n)+(2n^2)
-2m^2+2n^2-10m+5n-mn+8 is simplest form

You might be interested in

Use the slope to determine if lines AB and CD are parallel, perpendicular, or neither

ycow [4]

The slope of B is Q maybe

7 0

2 years ago

In a school of 330 students, 85 of them are in a drama club, 200 of them are in a sports team and 60 of them do both

Neko [114]

Answer:

105/330 = 21/66 = 7/22

7/22 is the answer I think

Step-by-step explanation:

n(U) = 330

n(A) = 85

n(B) = 200

n(A n B) = 60

Now,

n(A U B) = n(A) + n(B) - n(A n B)

or, n(A U B) = 85 + 200 - 60

or, n(A U B) = 285 - 60 = 225

Now,

n(A U B) compliment = n(U) - n(A U B)

n(A U B) compliment = 330 - 225

so, n(A U B) compliment = 105

Now,

probability = 105/330

8 0

1 year ago

I just need no. a please help me to prove this.

OleMash [197]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π and cos A = cos B · cos C

scratchwork:

A + B + C = π

A = π - (B + C)

cos A = cos [π - (B + C)] Apply cos

= - cos (B + C) Simplify

= -(cos B · cos C - sin B · sin C) Sum Identity

= sin B · sin C - cos B · cos C Simplify

cos B · cos C = sin B · sin C - cos B · cos C Substitution

2cos B · cos C = sin B · sin C Addition

$2=\dfrac{\sin B\cdot \sin C}{\cos B \cdot \cos C}$ Division

2 = tan B · tan C

$\text{Use the Sum Identity:}\quad \tan(B+C)=\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}$

<u>Proof LHS → RHS</u>

Given: A + B + C = π

Subtraction: A = π - (B + C)

Apply tan: tan A = tan(π - (B + C))

Simplify: = - tan (B + C)

$\text{Sum Identity:}\qquad \qquad \qquad =-\bigg(\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}\bigg)$

Substitution: = -(tan B + tan C)/(1 - 2)

Simplify: = -(tan B + tan C)/-1

= tan B + tan C

LHS = RHS: tan B + tan C = tan B + tan C $\checkmark$

5 0

2 years ago

Please help!!!

xxMikexx [17]

Answer:

A = 1.5 π in² ≈ 4.7 in²

Step-by-step explanation:

the area (A) of the sector is calculated as

A = area of circle × fraction of circle

= πr² × $\frac{\frac{\pi }{3} }{2\pi }$

= π × 3² × $\frac{1}{6}$

= 9π × $\frac{1}{6}$

= 1.5π in²

≈ 4.7 in² ( to the nearest tenth )

7 0

1 year ago

Which relation is a function?

aksik [14]

Answer:

Step-by-step explanation:

5 0

2 years ago