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2 years ago

Simplify the expression: 10s+–2s

Mathematics

2 answers:

Arada [10]2 years ago

5 0

Answer:

$\huge\boxed{8s}$

Step-by-step explanation:

Expression can be broken down into a couple of parts.

A term (like 10s, or -2s)
The variable (what the unknown value is being represented by, in this case, s)
The coefficient (the number next to the variable, showing what to multiply the variable by - in this case, 10 and -2)

When we have multiple terms with the same variable, we can perform operations on them assuming they have the same variable - for example, 5x + 6x = 11x, as 5+6=11.

We can apply the same logic here, however, note that adding a negative is the same as subtracting a positive. For example, 5 + -2 is the same as 5-2.

$10 - 2 = 8$

Therefore, our final expression is $8s$ .

Hope this helped!

likoan [24]2 years ago

3 0

Answer:

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Terms

Step-by-step explanation:

Step 1: Define

10s + -2s

Step 2: Simplify

Rewrite: 10s - 2s
Combine like terms: 8s

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3 years ago

Read 2 more answers

The doubling period of a bacterial population is 20 20 minutes. At time t = 100 t=100 minutes, the bacterial population was 9000

Nuetrik [128]

Answer:

The initial population was 2810

The bacterial population after 5 hours will be 92335548

Step-by-step explanation:

The bacterial population growth formula is:

$P = P_0 \times e^{rt}$

where P is the population after time t, $P_0$ is the starting population, i.e. when t = 0, r is the rate of growth in % and t is time in hours

Data: The doubling period of a bacterial population is 20 minutes (1/3 hour). Replacing this information in the formula we get:

$2 P_0 = P_0 \times e^{r 1/3}$

$2 = e^{r \; 1/3}$

$ln 2 = r \; 1/3$

$ln 2 \times 3 = r$

$2.08 \% = r$

Data: At time t = 100 minutes (5/3 hours), the bacterial population was 90000. Replacing this information in the formula we get:

$90000 = P_0 \times e^{2.08 \; 5/3}$

$\frac{9000}{e^{2.08 \; 5/3}} = P_0$

$2810 = P_0$

Data: the initial population got above and t = 5 hours. Replacing this information in the formula we get:

$P = 2810 \times e^{2.08 \; 5}$

$P = 92335548$

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2 years ago