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marshall27 [118]

3 years ago

11

Where is the point (2,-6) located in the coordinate plane? A. In quadrant 1 B. On the y-axis C. In quadrant III D. In quadrant I

V

1 answer:

ruslelena [56]3 years ago

5 0

Answer:

D. Quadrant IV

On the coordinate plane, you would go to the right two points, and then down 6, leading to quadrant IV

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The graph of which function has a minimum located at (4, –3)?

Oksanka [162]

It is the third option on Edge

5 0

3 years ago

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Zoologists are studying two newly discovered species of insects in a previously unexplored section of rain forest. They estimate

frosja888 [35]

Answer: Obtion B

$P = 1.3e ^ {-0.038t}\\\\P = 2.1e ^ {-0.046t}$

Step-by-step explanation:

The equation for exponential decay has the following form:

$y = pe ^{-rt}$

Where

p is the initial population

r is the rate of decrease

t is time.

In this problem we have to:

The current population of insect A to be 1.3 million and the current population of insect B to be 2.1 million.

So

$p_1 = 1.3$ in millions

$p_2 = 2.1$ in millions

We also need the populations of insect to be reduced at a rate of 3.8% and insect to be reduced at a rate of 4.6%.

so:

$r_1 = 0.038\\\\r_2 = 0.046$

then the exponential decay equation for insect A is:

$P = 1.3e ^ {-0.038t}$

the exponential decay equation for insect B is:

$P = 2.1e ^ {-0.046t}$

Finally, the system of equations is:

$P = 1.3e ^ {-0.038t}\\\\P = 2.1e ^ {-0.046t}$

The answer is the Option B

7 0

3 years ago

Please someone help I need help

azamat

Answer:

a b d are correct

Step-by-step explanation:

see the steps in the pic

7 0

4 years ago

Of 30 students 1/3 play sports of those students 2/5 play soccer, how many students play soccer

Natasha2012 [34]

4 students play soccer.

Rephrased:
2/3 of 1/3 of 30 total students play soccer.

Solution:
1/3×30=10
2/5×10=4

4 0

3 years ago

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If cosh x = 13 5 and x > 0, find the values of the other hyperbolic functions at x.

givi [52]

Recall that

$\cosh^2x-\sinh^2x=1\implies \cosh x=\pm\sqrt{1+\sinh^2x}$

You're given that $\cosh x=\dfrac{13}5>0$ , so omit the negative root. Then

$\dfrac{13}5=\sqrt{1+\sinh^2x}\implies\sinh^2x=\dfrac{144}{25}\implies\sinh x=\pm\dfrac{12}5$

Because $x>0$ , you have $\sinh x>0$ too, so omit the negative root again. Then $\sinh x=\dfrac{12}5$ .

Now,

$\mbox{sech }x=\dfrac1{\cosh x}=\dfrac5{13}$
$\mbox{csch }x=\dfrac1{\sinh x}=\dfrac5{12}$
$\tanh x=\dfrac{\sinh x}{\cosh x}=\dfrac{\frac{12}5}{\frac{13}5}=\dfrac{12}{13}$
$\coth x=\dfrac1{\tanh x}=\dfrac{13}{12}$

7 0

3 years ago