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gavmur [86]
3 years ago
10

A,B,C,D (need help ASAP)​

Mathematics
1 answer:
dsp733 years ago
3 0

Answer:

i think a

Step-by-step explanation:

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Can y'all help me with math I hate it
antoniya [11.8K]

Answer:

Step-by-step explanation:

the first one

6 0
3 years ago
Read 2 more answers
Could anyone help me on number 3?
xeze [42]

Answer:

n = 2m-4

Step-by-step explanation:

n+2/m-1=2

n+2=2(m-1)

n+2=2m-2

n=2m-2-2

n=2m-4

6 0
3 years ago
Marie made the box plot below to display a set of data. Which set of data could the box plot represent?
Assoli18 [71]

Answer:

  see below

Step-by-step explanation:

First of all, you want to find the data set the matches the extreme values of 5 and 35. That eliminates the 2nd and 4th choices.

Then you want to find the data set that has a median of 15. The first data set has a middle value (median) of 20, so that choice is eliminated.

The data set of the 3rd choice matches the box plot extremes, median, and quartile values.

7 0
3 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
In the problem below, AB, CD, and EF are two-digit numbers, where A, B, C, D, E and F represent distinct digits from 1 to 9. Is
Oksi-84 [34.3K]

Answer:

<em>When both the conditions hold true, F is prime.</em>

Step-by-step explanation:

AB, CD, and EF are two-digit numbers, where A, B, C, D, E and F represent distinct digits from 1 to 9.

  AB

+ CD

--------

  EF

1st condition, B and D are consecutive.

Adding B and D gives us F.

Possible values can be (F being the unit value after adding not considering the carry over):

B + D = F

1+2=3

2+3=5

3+4=7

4+5=9

5+6=1

6+7=3

7+8=5

8+9=7

Here F is not prime (because 9 is not prime).

Now, let us consider the 2nd condition as well.

i.e. C = 8

For the following

  AB

+ CD

--------

  EF

C is 8 then A must be 1 because any value other than 1 for A will make the sum of A and C greater than 9 and there will be a carry which is not the case here.

So, E = 8 + 1 = 9

Now, B  and D are consecutive and can not be 1, 8 or 9.

So, possible values are:

B + D = F

2 + 3 = 5

3 + 4 = 7

Here F is prime.

So, when both the conditions hold true, F is prime.

3 0
3 years ago
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