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4 years ago

What’s the rule for the transformation

Mathematics

2 answers:

Svet_ta [14]4 years ago

3 0

Answer:

the last one i think

Step-by-step explanation:

Alecsey [184]4 years ago

3 0

The answer to the question of yours is the first one

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What is the solution to this equation? 6x + 10 - 2x= 7 + 23

Strike441 [17]

Answer:

$\boxed{x=5}$

Step-by-step explanation:

First, you need to combine like terms in order to begin simplifying. This is done by subtracting 2x from 6x. Also, add 7 to 23.

6x + 10 - 2x = 7 + 23

4x + 10 = 7 + 23

4x + 10 = 30

Now, you have a two-step equation. Subtract 10 from both sides, and then divide by 4.

4x + 10 = 30

4x = 20

x = 5 is the final answer.

7 0

4 years ago

By the Triangle Inequality Theorem which set of side lengths could create a triangle?

Nonamiya [84]

Answer: B

Step-by-step explanation:

the sum of any two sides is bigger than the third side... so 5+9= 14, bigger than 6. 5+6=11, greater than 9. 9+6=15, bigger than 5.

4 0

3 years ago

Read 2 more answers

Scientific notation 100400

ad-work [718]

1.004 * 10^5
all scientific notation equations have to be a number greater than 1 but less than 10
then multiplied by 10 to the power of (move the decimal point to the right however many times you need, in this case 5) :)

5 0

4 years ago

Using the Addition Method, find the value of x in the

allochka39001 [22]

Step-by-step explanation:

3x+2y=2------------1

-3x+5y=5-----------2

adding 1 and 2 we get

7y=7

y=1

and from (1),we get

3x+2×1=2

3x=0

x=0 is the required value of x.

8 0

3 years ago

Please prove this........

Crazy boy [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

Proof LHS → RHS

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago