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4 years ago

Help please!!!!!!!!!!!

Mathematics

1 answer:

AysviL [449]4 years ago

7 0

Answer:

A. Plant 1 is growing at a faster rate than Plant 2.

Step-by-step explanation:

The slope (m) or rate of growth in Plant 1 is;

m = Change in y ÷ change in x

m = $\frac{3.5 - 1.5}{1 - 0}$ = 2

The linear equation of growth in Plant 1 is;

$\frac{y - 1.5}{x - 0}$ = 2

y = 2x + 1.5

The equation of growth in Plant 2 is; y = 1.5x + 3 (slope of 1.5)

Since the slope in growth of Plant 1 is greater than that in Plant 2, it implies then that Plant 1 is growing at a faster rate than Plant 2.

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Help!

lbvjy [14]

Answer:

Step-by-step explanatiWe know x is between 1 and 3 so the best estimate is 2 even though the answer could also be 3 but that will be skewed.

4 0

3 years ago

Read 2 more answers

Can anybody help me with this?

Alex787 [66]

Answer:

Step-by-step explanation:

The rate of change for the table is found using the slope formula. The rate of change for the linear function is the coefficient of x.

<h3>table</h3>

The slope formula tells us the rate of change is ...

m = (y2 -y1)/(x2 -x1)

m = (-9 -(-10))/(-1 -(-4)) = 1/3

<h3>equation</h3>

The coefficient of x in the equations of the offered answer choices are 1/4 and 6.

<h3>comparison</h3>

The relations between the various slopes are ...

1/4 < 1/3 < 6

That is, the equation with a slope of 1/4 has a rate of change less than that of the function represented by the table:

y = 1/4x +4 because ... table of values is 1/3

6 0

2 years ago

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

Alja [10]

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

<em><u>Solution:</u></em>

Given that,

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

<em><u>Use the following log rules</u></em>

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

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