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3 years ago

What is the reason for each step in the solution of the equation 12-5x+6x=4

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

7 0

Answer: x=-8

Step-by-step explanation: 12-5x+6x=4 1. Combine like terms: -5x+6x+12=4 2. Simplify: x+12-12=4-12 3. Subtract 4 from -12 so that you can find out what x is: x=-8

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What is the sum of series is numbers 64&66 ? Using the summation notation

iragen [17]

Answer:

64. 6138

66. 8.078125

Step-by-step explanation:

Let's take it one number at a time.

For number 64:

$6(2)^{1-1}=6$

$6(2)^{2-1}=12$

$6(2)^{3-1}=24$

$6(2)^{4-1}=48$

$6(2)^{5-1}=96$

$6(2)^{6-1}=192$

$6(2)^{7-1}=384$

$6(2)^{8-1}=768$

$6(2)^{9-1}=1536$

$6(2)^{10-1}=3072$

Sum = 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 + 1536 + 3072

Sum = 6138

For number 66:

$12(-\frac{1}{2})^{0}=12$

$12(-\frac{1}{2})^{1}=-6$

$12(-\frac{1}{2})^{2}=3$

$12(-\frac{1}{2})^{3}=-1.5$

$12(-\frac{1}{2})^{4}=0.75$

$12(-\frac{1}{2})^{5}=-0.375$

$12(-\frac{1}{2})^{6}=0.1875$

$12(-\frac{1}{2})^{7}=-0.0078125$

$12(-\frac{1}{2})^{8}=0.046875$

$12(-\frac{1}{2})^{9}=-0.0234375$

Sum = 12 + -6 + 3 + -1.5 + 0.75 + -0.375 + 0.1875 + -0.0078125 + 0.046875 + -0.0234375

or to make things simpler:

Sum = 12 - 6 + 3 - 1.5 + 0.75 - 0.375 + 0.1875 - 0.0078125 + 0.046875 - 0.0234375

Sum = 8.078125

4 0

3 years ago

A certain recipe requires 415 cups of flour and 237 cups of sugar. a) If 49 of the recipe is to be made, how much sugar is neede

Bas_tet [7]

49*237

11613

That's a lot of sugar, diabetes here we come :)

7 0

3 years ago

Which of the following is equivalent to (-2) -5? -2+(-5) -2+(-5) 2+(-5) 2-(-5)

gladu [14]

-2+(-5) is equal to (-2)-5.

5 0

3 years ago

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The area of a rectangle is 100 square

Nikitich [7]

Answer:

l = 100 / 8

Step-by-step explanation:

The standard formula for calculating area is bh = a.

Plug in 8 for our width.

8l = 100

Divide 8 on both sides.

l = 100 / 8.

7 0

2 years ago

A square has a side length 9 cm. What is its area?

Flauer [41]

Answer:30

Step-by-step explanation:

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3 years ago

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