All categories

3 years ago

13x−2= − 10 x − 2 −10x−2

Mathematics

1 answer:

Misha Larkins [42]3 years ago

3 0

Answer:

x=0

Step-by-step explanation:

13x-2 = 10x-2

Subtract 10x from each side

13x-10x-2 = 10x-2 -10x

3x-2 = -2

Add 2 for each side

3x-2+2 = -2+2

3x= 0

divide by 3

3x/3 = 0/3

x =0

You might be interested in

A survey showed that 70% of students would select roller coasters as their favorite ride at an amusement park. Out of 5,000 stud

Fiesta28 [93]

Answer:

3,500

Step-by-step explanation:

70% is equal to 0.70.

When you need to find a percent of a population, you multiply the decimal form to the population.

Therefore, with 70% of 5000, you multuiply 5000 by 0.70 which will give you 3500.

3,500 students would select roller coasters as their favorite ride.

7 0

3 years ago

Read 2 more answers

What is 5x5 answer for a cookie

34kurt

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Plz help me here I really need it. Please and thanks!

Alik [6]

Answer:

no they are not proportional and the graph would be at points (1, 3) (5, 9)

Step-by-step explanation:

Can i have brainliest Please?

8 0

2 years ago

What is 2 + 10? pls help!!!!!!

Jlenok [28]

12 because if you take 2 and add it to 10 you get 12

3 0

3 years ago

Read 2 more answers

Help ASAP show work please thanksss!!!!

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:

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

5 0

2 years ago