All categories

3 years ago

Which properties were used to solve the following problem?

Mathematics

1 answer:

Stels [109]3 years ago

5 0

Step-by-step explanation:

25(2x) + 25(3y)

25 · (2 · x) + 25 · (3 · y)

(25 · 2) · x + (25 · 3) · y Associative property (a · b) · c = a · (b · c)

(25 · 2)x + (25 · 3)y

(2 · 25)x + (3 · 25)y Commutative property a · b = b · a

50x + 75y

You might be interested in

Help will give brainly

Simora [160]

Answer:

Is it 72?

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the ratio a:b, if it is given that ... a+b=3a

grin007 [14]

The ratio of a : b is 1 : 2

Step-by-step explanation:

A ratio is a relationship between two numbers indicating how many times the first number contains the second number

Ex: If there are eight oranges and six bananas, then the ratio of oranges to bananas is eight to six, and is written as 8 : 6, which is equivalent to the ratio 4 :3 in its simplest form

∵ a + b = 3a

- To find the ratio between a and b find b in terms of a

- Subtract a from both sides

∴ b = 2a

→ a : b

Substitute b by 2a

→ a : 2a

Divide both terms by a

→ 1 : 2

∴ a : b = 1 : 2

The ratio of a : b is 1 : 2

Learn more:

You can learn more about the ratio in brainly.com/question/4302397

#LearnwithBrainly

5 0

4 years ago

Plz help... I’m offering 10 pts

madam [21]

Answer:

B._________________________

6 0

3 years ago

Simplify -7 1/8 - -9 1/2. (10 points)

Cerrena [4.2K]

$2 \frac{3}{8}$

3 0

4 years ago

Find the arc length of the given curve between the specified points.

yuradex [85]

Answer:

4.25

Step-by-step explanation:

Given:

$f(x) = \frac{x^3}{12} + \frac{1}{x} \\\\Arc Length = \int\limits^a_b {\sqrt{1 + (f'(x))^2} } \, dx \\f '(x) = \frac{x^2}{6} - \frac{1}{x^2}\\\\f'(x)^2 = (\frac{x^2}{6} - \frac{1}{x^2})^2 = \frac{x^4}{36} + \frac{1}{x^4} - \frac{1}{3}\\\\1 + f'(x)^2 = \frac{x^4}{36} + \frac{1}{x^4} + \frac{2}{3}\\\\\= \frac{x^8 + 36 + 12x^4}{36x^4}\\\\= \frac{(x^4 + 6)^2}{36x^4}\\\\=\sqrt{1 + f'(x)^2} = \sqrt{ \frac{(x^4 + 6)^2}{36x^4}}\\\\= \frac{x^2}{6} + \frac{1}{x^2} \\\\$

$ArcLength = \int\limits^4_1 {\frac{x^2}{6} + \frac{1}{x^2} } \, dx \\= (\frac{x^3}{18} - \frac{1}{x})\limits^4_1\\\\= (\frac{64}{18} - \frac{1}{4}) - (\frac{1}{18} - 1)\\\\= \frac{17}{4}= 4.25$

8 0

3 years ago