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tensa zangetsu [6.8K]
3 years ago
15

A veterinarian will prescribe an antibiotic to a dog based on its weight. The effective dosage of the antibiotic is given by d ≥

1∕5w2, where d is dosage in milligrams and w is the dog's weight in pounds. Which of the following ordered pairs gives an effective dosage of antibiotics for a 35-pound dog?
Mathematics
1 answer:
gregori [183]3 years ago
6 0

Answer:

B) (35, 260)

Step-by-step explanation:

A veterinarian will prescribe an antibiotic to a dog based on its weight. The effective dosage of the antibiotic is given by d ≥ 1∕5w2, where d is dosage in milligrams and w is the dog's weight in pounds. Which of the following ordered pairs gives an effective dosage of antibiotics for a 35-pound dog?

A) (35, 240)

B) (35, 260)

C) (260, 35)

D) (240, 35)

Ordered pairs is composed of pairs, usually an x coordinate and a y coordinate. It refers to a location of a point on the coordinate. It matches numbers to functions or relations.

Given the relation between d is dosage in milligrams and w is the dog's weight in pounds as d ≥ 1∕5w²

For a 35 pound dog (i.e w = 35 pound). The dosage is given as:

d ≥ 1∕5(35)² ≥ 245 milligrams.

For an ordered pair (x, y), x is the independent variable (input) and y is the dependent variable (output).

The dog weight is the independent variable and the dosage is the dependent variable.

From the ordered pairs, the best option is (35, 260) because 260 ≥ 240

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Alchen [17]
Ooh, fun

geometric sequences can be represented as
a_n=a(r)^{n-1}
so the first 3 terms are
a_1=a
a_2=a(r)
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the sum is -7/10
\frac{-7}{10}=a+ar+ar^2
and their product is -1/125
\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3

from the 2nd equation we can take the cube root of both sides to get
\frac{-1}{5}=ar
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r
subsituting -1/5 for ar
\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r
which simplifies to
\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for ax^2+bx+c=0
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}
so
for 2r²-5r+2=0
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r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}
r=\frac{5 \pm \sqrt{25-16}}{4}
r=\frac{5 \pm \sqrt{9}}{4}
r=\frac{5 \pm 3}{4}
so
r=\frac{5+3}{4}=\frac{8}{4}=2 or r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}

use them to solve for the value of a
\frac{-1}{5}=ar
\frac{-1}{5r}=a
try for r=2 and 1/2
a=\frac{-1}{10} or a=\frac{-2}{5}


test each
for a=-1/10 and r=2
a+ar+ar²=\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}
it works

for a=-2/5 and r=1/2
a+ar+ar²=\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}
it works


both have the same terms but one is simplified

the 3 numbers are \frac{-2}{5}, \frac{-1}{5}, and \frac{-1}{10}
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Answer:

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