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Yakvenalex [24]

3 years ago

11

In 1 through 3, what is the relationship between the values of the given digits?

1 answer:

quester [9]3 years ago

6 0

Answer:

7000 (7 thousand)

700 (7 hundred)

20 (2 tens)

2 (2 units)

Step-by-step explanation:

what is the relationship between the values of the given digits?

1. The 7s in 7,700

2. The 2's in 522

From the knowledge of place values;

7,700 could be broken down thus :

7000 + 700 + 0 + 0

The first 7 depicts thousands as it has 3 trailing digits (7000)

The second 7 depicts hundred as it has 2 trailing digits (700)

522 could be broken down thus :

500 + 20 + 2

From 522

The first '2' has one trailing digit = tens

The ending / last digit ia always = Unit value

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Evaluate the six trigonometric functions of the angle 0

Alla [95]

# Sin θ = 9/15.

# Cos θ = 12/15.

# Cosec θ = 15/9.

# Sec θ = 5/4

Step-by-step explanation:

$by \: using \: pythagorian \: triplets$

${a}^{2} + {b}^{2} = {c}^{2}$

${9}^{2} + {12}^{2} = {c}^{2}$

$81 + 144 = {c}^{2}$

$225 = {c}^{2}$

$c = 15.$

$\sin θ = \frac{9}{15}$

$\cos θ = \frac{12}{15} = \frac{4}{5}$

$\csc θ = \frac{15}{9}$

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irga5000 [103]

Answer:

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Logs are stacked in a pile. The bottom row has 50 logs and next to bottom row has 49 logs. Each row has one less log than the ro

Mars2501 [29]

Answer:

46 logs on the 5th row.

Step-by-step explanation:

Number of logs on the nth row is

n = 50 - (n-1)

n = 51 - n (so on the first row we have 51 - 1 = 50 logs).

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Wewaii [24]

Answer:

1. $P(x)$ ÷ $Q(x)$ ---> $\frac{-3x + 2}{3(3x - 1)}$

2. $P(x) + Q(x)$ ---> $\frac{2(6x - 1)}{(3x - 1)(-3x + 2)}$

3. $P(x) - Q(x)$ ---> $\frac{-2(12x - 5)}{(3x - 1)(-3x + 2)}$

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Step-by-step explanation:

Given that:

1. $P(x) = \frac{2}{3x - 1}$

$Q(x) = \frac{6}{-3x + 2}$

Thus,

$P(x)$ ÷ $Q(x)$ = $\frac{2}{3x - 1}$ ÷ $\frac{6}{-3x + 2}$

Flip the 2nd function, Q(x), upside down to change the process to multiplication.

$\frac{2}{3x - 1}*\frac{-3x + 2}{6}$

$\frac{2(-3x + 2)}{6(3x - 1)}$

$= \frac{-3x + 2}{3(3x - 1)}$

2. $P(x) + Q(x)$ = $\frac{2}{3x - 1} + \frac{6}{-3x + 2}$

Make both expressions as a single fraction by finding, the common denominator, divide the common denominator by each denominator, and then multiply by the numerator. You'd have the following below:

$\frac{2(-3x + 2) + 6(3x - 1)}{(3x - 1)(-3x + 2)}$

$\frac{-6x + 4 + 18x - 6}{(3x - 1)(-3x + 2)}$

$\frac{-6x + 18x + 4 - 6}{(3x - 1)(-3x + 2)}$

$\frac{12x - 2}{(3x - 1)(-3x + 2)}$

$= \frac{2(6x - 1}{(3x - 1)(-3x + 2)}$

3. $P(x) - Q(x)$ = $\frac{2}{3x - 1} - \frac{6}{-3x + 2}$

$\frac{2(-3x + 2) - 6(3x - 1)}{(3x - 1)(-3x + 2)}$

$\frac{-6x + 4 - 18x + 6}{(3x - 1)(-3x + 2)}$

$\frac{-6x - 18x + 4 + 6}{(3x - 1)(-3x + 2)}$

$\frac{-24x + 10}{(3x - 1)(-3x + 2)}$

$= \frac{-2(12x - 5}{(3x - 1)(-3x + 2)}$

4. $P(x)*Q(x) = \frac{2}{3x - 1}* \frac{6}{-3x + 2}$

$P(x)*Q(x) = \frac{2*6}{(3x - 1)(-3x + 2)}$

$P(x)*Q(x) = \frac{12}{(3x - 1)(-3x + 2)}$

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