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Strike441 [17]
3 years ago
6

you have dance class every 3 days today is monday and you have dance class in how many more days will you you have dance class o

n monday again use the lcm to find your answer
Mathematics
2 answers:
Cerrena [4.2K]3 years ago
4 0
7 days in a week, 3 days a rotation.
multiples of 7: 7, 14, 21, 28
multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24

the LCM is 21, because it is the smallest multiple of both numbers. there will be 21 more days until you take dance class on mondays.
Tpy6a [65]3 years ago
3 0
The awnser would be : Lmc 21
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Answer:

Excuse me I need more information to solve this question.

Step-by-step explanation:

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3 years ago
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What is the area of a parallelogram whose vertices are A(−12, 2) , B(6, 2) , C(−2, −3) , and D(−20, −3) ? Enter your answer in t
agasfer [191]

Answer:

  \boxed{90}\,\text{units$^2$}

Step-by-step explanation:

The sides AB and CD are parallel to the x-axis and 5 units apart. The length of side AB is 6-(-12) = 18 units. The area is the product of these dimensions:

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4 0
4 years ago
I NEED HELP PLEASE ?!!
Fittoniya [83]
The circle equation is in the format (x – h)² + (y – k)² = r², with the center being at the point (h, k) and the radius being "r".

QUESTION 11.
Equation x²+y²+10x-14y-7 =0 can be rewritten as: x²+10x+25 + y² -14y + 49 -7 - 25 - 49=0
It can be factories as (x + 5)² + (y – 7)² = 9²

Therefore the radius equals 9 and the center is (-5,7)

QUESTION 12.
From equation (x + 4)² + y² = 121

The radius equals √121 = 11 and the center is (-4,0)

QUESTION 13.
As there are missing information in the question, I can't assist. However, you can use the general circle equation (x – h)² + (y – k)² = r² to solve the question.

Finally equations 14 & 15 aren't linear.

Hope that helps you :)
5 0
3 years ago
Can some one please help me
Nady [450]
40 = 1 x 2 x 2 x 2 x 5
8 0
4 years 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
3 years ago
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