Test: PRACTICE EXAM #1

1 answer:

4 0

John should receive about 7200 milligrams of the drug.The unit conversion given is,1 lb =0.45 kg.

<h3>What is weight?</h3>

The weight of matter is found as the product of the mass and the gravitational acceleration.

Unit conversion;

1 lb =0.45 kg.

Dosage of a drug = 200 mg / kilogram

John weighs = 80 lb

1 lb =0.45 kg.

80 lb = 36 kg

According to the weight the drug should he receive is;

Dosage of a drug = 200 mg / kilogram × 36

Dosage of a drug = 7200 milli gram

Hence, John should receive about 7200 milligrams of the drug.

To learn more about the weight refer;

brainly.com/question/10069252

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$e)\\\\x^2-7x+10=0\\\\x^2-7x=-10\\\\x^2-7x+(\frac72)^2=-10+(\frac72)^2\\\\x^2-7x+(\frac72)^2=-10+(\frac72)^2\\\\(x-\frac72)^2=-\frac{20}4+\frac{49}4\\\\ (x-\frac72)^2=\frac{29}4\\\\x-\frac72=\pm\sqrt{\frac{29}4}\\\\ \underline{x=\pm\frac{\sqrt{29}}2+\frac72\approx6.19\ or\ 0.81}\\\\\\f)\\\\x^2+5x-2=0\\\\x^2+5x=2\\\\x^2+5x+(\frac52)^2=2+(\frac52)^2\\\\(x+\frac52)^2=2+6.25 \\\\x+\frac52=\pm\sqrt{8.25}\\\\\underline{x=\pm\sqrt{8.25}-\frac52\approx-5.37\ or\ 0.37}$

$g)\\\\x^2+9x-4=0\\\\x^2+9x=4\\\\x^2+9x+(\frac92)^2=4+(\frac92)^2\\\\(x+\frac92)^2=4+20.25\\\\x+\frac92=\pm\sqrt{24.25}\\\\\underline{x=\pm\sqrt{24.25}-4.5\approx-9.42\ or\ 0.52}$

$h)\\\\x^2-3x-6=0\\\\x^2-3x=6\\\\x^2-3x+(\frac32)^2=6+(\frac32)^2\\\\(x-\frac32)&^2=6+2.25\\\\x-\frac32=\pm\sqrt{8.25}\\\\ \underline{x=\pm\sqrt{8.25}+1.5\approx4.37\ or-1.37}$