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4 years ago

Can some body please help meeeee T^T

Mathematics

2 answers:

Nata [24]4 years ago

3 0

Answer:

Till tomorrow

TT Definition / TT Means

The definition of TT is "Telegrahic Transfer" or "Till tomorrow"

Step-by-step explanation:

pav-90 [236]4 years ago

3 0

Answer:

The answer is 5

Step-by-step explanation:

(50-10)=40/8=5

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Alejandra tiene una cubeta de 1 1/2 litros de pintura que le sobraron de su último mural. Para no cargar toda la cubeta, ella di

laiz [17]

Answer:

The correct answer is 6 jars can be used

Step-by-step explanation:

According to the given scenario, the calculation of the number of jars needed is shown below:

The 1 1 ÷ 2 liters means = 3 ÷ 2 liters

And, the paint is divided into 1 ÷4 liter jars

So, the number of jars needed is

= (3 ÷ 2) ÷ ( 1 ÷ 4)

= (4 × 3) ÷ 2

= 6 jars needed

Therefore the number of jars needed is 6 jars

4 0

3 years ago

Please help me to prove this!

Sophie [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

Proof LHS → RHS:

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

$\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]$

$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$