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Sophie [7]
4 years ago
6

a certain number was increased by 5 then decreased by 3 than halved the result was 0.3 less than the original number

Mathematics
2 answers:
SCORPION-xisa [38]4 years ago
6 0

Answer:

\large \boxed{2.6}

Step-by-step explanation:

         Let x = the number

Then x + 5 = the number increased by 5

    x + 5 - 3 = then decreased by 3

½(x + 5 - 3) = then halved

      x - 0.3 = 0.3 less than the number

\begin{array}{rcl}\frac{1}{2}(x + 5 - 3) & = & x - 0.3\\\frac{1}{2}(x + 2) & = & x - 0.3\\x + 2 & = & 2x - 0.6\\x + 2.6 & = & 2x\\x & = &\mathbf{2.6}\\\end{array}\\\text{The number is $\large \boxed{\mathbf{2.6}}$}

Check:

½(2.6 + 5 - 3) = 2.6 - 0.3

          ½(4.6) = 2.3

               2.3 = 2.3

OK.

mina [271]4 years ago
5 0

Answer: 20

Step-by-step explanation:

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slamgirl [31]

Answer:Solve. 2a + 3b = 5 6= a -5 a = 4 b = -1 1 a = 6 b=1 a=6 6 = -1 a=4 6 = 1

Step-by-step explanation:

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3 years ago
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(X- 3)² - 0

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8 0
2 years ago
the mass of a sphere made of gold is proportional to the diameter, d, cubed. express the mass m of the sphere as a function of t
elixir [45]

Answer:

m         =   cd³

Step-by-step explanation:

mass of        is proportional to          the diam

sphere                                                 cubed

Symbolically, this can be written as:

  m         =   cd³

4 0
3 years ago
Account is overdrawn by $50. You write a check for $20. What is the balance in your account?
Ad libitum [116K]

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30

Step-by-step explanation:

3 0
3 years ago
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Please help me to prove this!​
Sophie [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

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)

<u>Proof LHS → RHS:</u>

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)

\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C   \checkmark

5 0
3 years ago
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