Mathematics, 05.09.2020 01:01 Crxymia

Find the area of the triangle with vertices: Q(4,4,-1), R(5,7,-2), S(3,2,-4).

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Answer from: MayFlowers

The area of the triangle is 5.875 square units.

Step-by-step explanation:

First, we need to calculate the lengths of the triangle by means of Pythagorean Theorem:

$a = \|\overrightarrow {QR}\| = \sqrt{(5-4)^{2}+(7-4)^{2}+(-2+1)^{2}}$

$a \approx 3.317$

$b = \|\overrightarrow {RS}\| = \sqrt{(3-5)^{2}+(2-7)^{2}+(-4+2)^{2}}$

$b \approx 5.745$

$c = \|\overrightarrow {QS}\| = \sqrt{(3-4)^{2}+(2-4)^{2}+(-4+1)^{2}}$

$c \approx 3.742$

Now, we can use the following formula to fid the area of the triangle (), measured in square units:

$A = \sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}$

Where:

- Semiperimeter of the triangle, dimensionless.

, , - Sides of the triangle, dimensionless.

The semiperimeter can be determined by this:

$s = \frac{a+b+c}{2}$

If we know that $a \approx 3.317$ , $b \approx 5.745$ and $c \approx 3.742$ , the area of the triangle is:

$s = \frac{3.317+5.745+3.742}{2}$

s = 6.402

$A = \sqrt{(6.402)\cdot (6.402-3.317)\cdot (6.402-5.745)\cdot (6.402-3.742)}$

$A \approx 5.875$

The area of the triangle is 5.875 square units.

Answer from: Quest

)because it's going in order of the math sign

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

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Find the area of the triangle with vertices: Q(4,4,-1), R(5,7,-2), S(3,2,-4)....