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r-ruslan [8.4K]
2 years ago
11

Help!!

Mathematics
1 answer:
Elza [17]2 years ago
3 0

Answer:

(b)\ 2x - 9 = -1

(c)\ -2x + 12 = 4

(e)\ 2(x - 10) = -12

Step-by-step explanation:

Required

Which equals x = 4

(a)\ 2x + 6 = 2

Collect like terms

2x  = 2 - 6

2x  = -4

Divide both sides by 2

x = -2

(b)\ 2x - 9 = -1

Collect like terms

2x = -1 + 9

2x = 8

Divide both sides by 2

x = 4

(c)\ -2x + 12 = 4

Collect like terms

-2x =- 12 + 4

-2x =-8

Divide both sides by -2

x = 4

(d)\ 2(x + 2) = 10

Divide both sides by 2

x + 2 =5

Collect like terms

x = 5-2

x = 3

(e)\ 2(x - 10) = -12

Divide both sides by 2

x - 10 = -6

Collect like terms

x = 10 - 6

x = 4

Hence, the equations with the required solution are:

(b)\ 2x - 9 = -1

(c)\ -2x + 12 = 4

(e)\ 2(x - 10) = -12

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Answer: See explanation

Step-by-step explanation:

You didn't give the options relating to the question.

To solve this first, you need to know that we have to note that (+) × (-) = (-). Therefore, 1.25+(-0.75) will be thesame as writing 1.25 - 0.75. This will then be:

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Answer: 27/36

Step-by-step explanation: It says there are 36 bikes (18 pairs), meaning that there are 2 bikes per pair because 18 multiplied by 2 is 36.

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Llana [10]

Answer:

\displaystyle log_\frac{1}{2}(64)=-6

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

log_b(1) = 0

Logarithm of the base:

log_b(b) = 1

Product rule:

log_b(xy) = log_b(x) + log_b(y)

Division rule:

\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)

Power rule:

log_b(x^n) = n\cdot log_b(x)

Change of base:

\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

\displaystyle log_\frac{1}{2}(64)

Factoring 64=2^6.

\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)

Since

\displaystyle 2=(1/2)^{-1}

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})

Applying the logarithm of the base:

\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}

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