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3 years ago

On a line graph, time is usually represented on the vertical axis.

1 answer:

3 0

False. Time is usually represented on the horizontal axis.

This is because the independent variable usually goes on the horizontal/x axis and the dependent variable goes on the vertical/y axis. The independent variable is what you're changing and usually time is what affects whatever outcome (aka dependent variable value) you get.

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How many real roots and how many complex roots exist for the polynomial

Volgvan

Answer:

D. 4 real roots and 0 complex roots

Step-by-step explanation:

If I assume that the function you are saying is

$F(x)=x^4+x^3-5x^2+x-6$

There should be up to "4 roots," there can't be more or less than 4 total solutions. First, we need to check how many sign changes are there in this function. There are 3 positive real roots. Now lets check for negative roots.

$F(-x)=x^4-x^3-5x^2-x-6$

There are is only 1 negative real root. Since we basically have 4 real roots, and the max is 4. There should be 4 real roots and 0 complex roots.

8 0

4 years ago

HELP ME WITH THIS PROBLEM ASAP!!!

Oliga [24]

Answer: B

Step-by-step explanation:

For the left side of the graph, the dot is not shaded so it is not equal, thus a and d are wrong.

c is wrong as well as the left side of the graph is x less than 1 and not more than so from the process of elimination, we get that the answer is B

6 0

2 years ago

2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA

professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

$A=45e^{-0.0045(t)}$

a) To find the initial amount of this substance

At t=0, we get

$A=45e^{-0.0045(0)}$ $A=45e^0$

We know that e^0=1 ( anything to the power zero is 1)

we get,

$A=45$

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

$\frac{45}{2}=45e^{-0.0045(t)}$

$\frac{1}{2}=e^{-0.0045(t)}$

Taking natural logarithm on both sides we get,

$\ln (\frac{1}{2})=-0.0045(t)^{}$ $(-1)\ln (\frac{1}{2})=0.0045(t)$ $\ln (\frac{1}{2})^{-1}=0.0045(t)$ $\ln (2)=0.0045(t)$ $0.6931=0.0045(t)$ $t=\frac{0.6931}{0.0045}$ $t=154.02$

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,