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3 years ago

Which is the largest interval over which the function shown in the table is increasing?

Mathematics

1 answer:

vova2212 [387]3 years ago

3 0

Answer:

x < 0

Step-by-step explanation:

If you were to graph the function, you would get a parabola that opens down with the vertex at (0, 0). So, the graph is increasing from -∞ to 0.

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Does this expression represent the sum 13.76 + 2.8? 14 + 4/100

aleksklad [387]

Answer:

This might be wrong.

False?

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3 years ago

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What are all of the real roots of the following polynomial? f(x) = x4 - x3 - 13x2 + x + 12

leonid [27]

We can find roots by setting f(x) = 0 --> x^4 - x^3 - 13x^2 + x + 12 = 0.

Factoring f(x) to make things easier for us gives us (x - 1)(x + 1)(x - 4)(x + 3), which, equal to 0, gives us the ability to set each set of parentheses = 0.

x - 1 = 0 gives us x = 1.

x + 1 = 0 --> x = -1.

x - 4 = 0 --> x = 4.

x + 3 = 0 --> x = -3.

our roots are x = 1, -1, 4, and -3.

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3 years ago

I am bad at math, and don't understand this... √(7x+1) - 4 = 11

Snowcat [4.5K]

Answer:

Step-by-step explanation:

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3 years ago

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Bt= 8500 *(8/27)^t/3After a special medicine is introduced into a Petri dish full of bacteria, the number of bacteria remaining

ozzi

Answer:

Every 1.71 seconds, the bacteria loses $\frac{1}{2}$

Step-by-step explanation:

Given

$B(t) = 8500 * (\frac{8}{27})^\frac{t}{3}\\$

Required [Missing from the question]

Every __ seconds, the bacteria loses $\frac{1}{2}$

First, we model the function from t/3 to t.

$B(t) = 8500 * (\frac{8}{27})^\frac{t}{3}\\$

Apply law of indices

$B(t) = 8500 * (\frac{8^\frac{1}{3}}{27^\frac{1}{3}})^t$

Evaluate each exponent

$B(t) = 8500 * (\frac{2}{3})^t$ --- This gives the number of bacteria at time t

At time 0, we have:

$B(0) = 8500 * (\frac{2}{3})^0$

$B(0) = 8500 * 1$

$B(0) = 8500$

Let r be the time 1/2 disappears.

When 1/2 disappears, we have:

$B(r) = \frac{B(0)}{2}$

$B(r) = \frac{8500}{2}$

$B(r) = 4250$

So, we have:

$B(t) = 8500 * (\frac{2}{3})^t$

Substitute r for t

$B(r) = 8500 * (\frac{2}{3})^r$

Substitute $B(r) = 4250$

$4250 = 8500 * (\frac{2}{3})^r$

Divide both sides by 8500

$\frac{4250}{8500} = (\frac{2}{3})^r$

$\frac{1}{2} = (\frac{2}{3})^r$

Take log of both sides

$log(\frac{1}{2}) = log (\frac{2}{3})^r$

Apply law of logarithm

$log(\frac{1}{2}) = r\ log (\frac{2}{3})$

Make r the subject

$r = log(\frac{1}{2}) / log (\frac{2}{3})$

$r = \frac{-0.3010}{-0.1761}$

$r = 1.71$

Hence, it reduces by 1/2 after every 1.71 seconds

6 0

3 years ago

Chase took a taxi from his house to the airport the taxi company charged a pick up fee $1.20 plus $4.75 Per mile . The total far

taurus [48]

I believe that the taxi went 10 miles

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3 years ago