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3 years ago

Help me plzz help me plzz pleade

Mathematics

1 answer:

kirill [66]3 years ago

6 0

Answer:

Step-by-step explanation:

it is four because count both sides

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9d-4d-2d+8=-3d help pls

goblinko [34]

Answer:

-8/3 = d OR -2 2/3

Step-by-step explanation:

9d-4d-2d+8=-3d Well first combine all like terms

3d+8=-3d Now isolate the variable by subtracting 4d from both sides

3d(-3d) + 8= -3d (-3d)

8= -3d Divide both sides by -3

-8/3 = d OR -2 2/3

6 0

3 years ago

When two six-sided dice are rolled what is the probability that the product of their scores will be greater than six?

Pachacha [2.7K]

Answer: $\bold{\dfrac{11}{18}}$

Step-by-step explanation:

Think of the products row by row:

11 12 13 14 15 16 - 0 products greater than 6

21 22 23 24 25 26 - 3 products greater than 6

31 32 33 34 35 36 - 4 products greater than 6

41 42 43 44 45 46 - 5 products greater than 6

51 52 53 54 55 56 - 5 products greater than 6

61 62 63 64 65 66 - 5 products greater than 6

$\dfrac{\text{number greater than 6}}{\text{total possible outcomes}}=\dfrac{22}{36}=\dfrac{11}{18}\ when\ reduced$

8 0

3 years ago

Read 2 more answers

Determine which of the following is a rectangle form of a complex number.

Varvara68 [4.7K]

Answer:

A) a+bi

Explanation:

Rectangular form of complex numbers is written as a+bi, where a and b are integers.

Rectangular form always has an integer piece, a, and a complex piece, bi.

4 0

3 years ago

Westkost [7]

Answer:

he would do 9 weeks of school

Step-by-step explanation:

count by fives

3 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

Help me plzz help me plzz pleade​

Help me plzz help me plzz pleade