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3 years ago

14

Dr. Hung prescribed 0.019 L more medicine then Dr. tannenbaum Dr. Evans prescribe 0.02 was the one Doctor Who prescribed the mos

t who prescribed the lease

1 answer:

arlik [135]3 years ago

4 0

Dr. Evans prescribed more. Dr. Hung prescribed less. 0.019 < 0.02

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1. What is the difference between 6 and 3<br> A. 6+3<br> B. 6-3<br> C. 6/3<br> D. 6x3

antiseptic1488 [7]

Answer:

B. 6-3

Step-by-step explanation:

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3 years ago

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Can you please help me solve this?

oee [108]

So, i would use the add the two equation method, so add those to get:
2e=-4
e=-2
then plug in e to get d, so d+-2=1
d=3
plug into the other one to make sure
-3+-2=-5
so it would be (3,-2)

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3 years ago

I GIVE BRAINLILSTE<br> .................................

Archy [21]

The answer is a or d i’m not sure which one though sorry.

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2 years ago

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Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

a)

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

b)

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

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10 months ago

Why must one leg of a triangle that the Pythagorean triple represents need an even-numbered length?

pashok25 [27]

Most of the square numbers are even.so if we need to find root over we need an even number to convert it to perfect square.

See the example below

P=3
B=4
H=5

$\\ \ast\bull\rm\longmapsto H^2=P^2+B^2$

$\\ \ast\bull\rm\longmapsto H^2=3^2+4^2$

$\\ \ast\bull\rm\longmapsto H^2=9+16$

$\\ \ast\bull\rm\longmapsto H^2=25$

$\\ \ast\bull\rm\longmapsto H=\sqrt{25}$

$\\ \ast\bull\rm\longmapsto H=5$

Some other triplets are

(6,8,10)
(5,12,13)
(8,15,17)

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2 years ago