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3 years ago

Is the product of two integers positive, negative , or zero ? How can you tell?

Mathematics

1 answer:

Firlakuza [10]3 years ago

6 0

Answer:it depends on the numbers you are using.

Step-by-step explanation:

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In the table below, x represents the miles traveled and y represents the cost to travel by train.

Afina-wow [57]

Answer:

Third option

m = 2.25

Step-by-step explanation:

The slope of the function is equal to its rate of change between two points.

For example, for a couple of points $(x_1, y_1),\ (x_2, y_2)$

The slope is:

$m =\frac{y_2-y_1}{x_2-x_1}$

In this problem we have the following points

1.(2, 8.50) , 2. (5, 15.25), 3. (8, 22.00) , 4. (12, 31.00)

We calculate the slope between the first two points

$m =\frac{15.25-8.50}{5-2}=2.25$

We calculate the slope between the points 3 and 4

$m =\frac{31-22}{8-4}=2.25$

We calculate the slope between the points 2 and 3

$m =\frac{22-15.25}{8-5}=2.25$

The slope is always the same. Therefore m is constant and the fution is a line of slope m = 2.25

7 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST 2 hours x 47 mph please show work for brainliest

gtnhenbr [62]

Answer:

2 hours * (47 miles / hour) = 94 miles

multiply the value in miles per hour by the conversion factor '1.609344'.

So, 2.2 miles per hour = 2.2 × 1.609344 = 3.5405568 kilometers per hour.

Step-by-step explanation:

7 0

3 years ago

If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

Solution:

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

Use the following log rules

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

$b) log_b 324$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

$c) log_b\frac{8}{9}$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

$d) \frac{log_b27}{log_b16}$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$