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UNO [17]
3 years ago
9

Hii I just wasn't sure about how to answer this so could someone help?

Mathematics
2 answers:
balandron [24]3 years ago
4 0

Answer:

1/2

Step-by-step explanation:

AlladinOne [14]3 years ago
4 0

Answer:

1/9

Step-by-step explanation:

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2/3 (6y - 4) = - 4y <br>​
Nookie1986 [14]

Answer:

y = 1/3

Step-by-step explanation:

2/3 (6y - 4) = - 4y

2(6y-4)/3 = -4y

2(6y-4) = (-3)(4y)

y = 1/3

7 0
2 years ago
Read 2 more answers
Which are correct representations of the inequality 6x2 3 + 4(2x - 1)? Select three options.
olga2289 [7]

Answer:

Option A, B, and C.

Step-by-step explanation:

Given, 6x ≥ 3 + 4(2x - 1),

Solve to find the correct representations given in the options.

6x ≥ 3 + 4(2x - 1)

Apply distributive property

6x ≥ 3 + 8x - 4 (option B is correct) ✅

Add like terms

6x ≥ -1 + 8x

Add 1 to both sides

6x + 1 ≥ 8x

Subtract 6x from each side

1 ≥ 8x - 6x

1 ≥ 2x (option A is correct) ✅

Divide both sides by 2

1/2 ≥ 2x/2

½ ≥ x

½ ≥ x means all possible values of x are less than 0.5. representing this inequality on a graph, we would have the directed line starting at 0.5 moving towards our left.

This make option C correct.✅

5 0
2 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
2 years ago
Need help don’t know what I’m doing
docker41 [41]
This problem mean you have to find the point that the Domain and Range is suit (I'm sorry English is my second language) 
Example:
1) G ; 2) Y
If you want me to do the rest of it tell me Okay ^^

8 0
2 years ago
The following table gives the science scores and times spent studying for 10 students:
AveGali [126]
There appears to be a positive correlation between the number of hour spent studydng and the score on the test.

When identifying the independent and dependent quantities, we think about what would cause the other to change.  The score on the test would not cause the number of hours spent studying to change; rather, the number of hours spent studying would cause the score to change.  This means that the number of hours studying would be the independent quantity and the score would be the dependent quantity.

Plotting the graph with the time studying on the x-axis (independent) and the score on the y-axis (dependent) gives you the graph shown.  You can see in the image that there seems to be a positive correlation; the data seem to generally be heading upward.

8 0
3 years ago
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