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3 years ago

15

Trent walks 133 yards due south, then 168 yards due west and finally 293 yards due north. How far is Trent from his starting poi

nt?

1 answer:

Deffense [45]3 years ago

4 0

Answer:

164 yards

Step-by-step explanation:

take 293 minus 133 and get 160 then take 168

then add them together then divide by 2

You will get:

do not forget your units

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A package of 6 pairs of insulated gloves costs $51.54. What is the unit price of the pairs of gloves ?

mafiozo [28]

Answer:

$8.59

Step-by-step explanation:

The unit price is just asking for what one pair of gloves cost.

$\$ 51.54 \div6=\$8.59$

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2 years ago

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there isnt a list of options given but should be something like

38 + 35 + 22 = 22 + 35 + 38

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2 years ago

What is the true solution to In 20+ In 5= 2 In x?<br> x=5. A<br> X= 10 b<br> X=50 c<br> X= 100 d

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ln(20) + ln(5) = 2 ln(<em>x</em>)

ln(20×5) = ln(<em>x</em> ²)

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100 = <em>x</em> ²

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3 years ago

There are 20 students in a math class. Thirty percent of the

Mekhanik [1.2K]

Answer:

Step-by-step explanation:

20% of 30 is 6 are randomly selected.

30 choose 6 = 30! / (30-6)! 6!

= 30! / 24!6!

= 30 * 29 * 28 * 27 * 26 * 25

--------------------------------- <-- 24! cancels

6 * 5 * 4 * 3 * 2 * 1

= 29 * 28 * 27 * 26 * 25

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2 years ago

When a bactericide is added to a nutrient broth in which bacteria are growing, the bacteria population continues to grow for a

baherus [9]

Answer:

a) 1296 bacteria per hour

b) 0 bacteria per hour

c) -1296 bacteria per hour

Step-by-step explanation:

We are given the following information in the question:

The size of the population at time t is given by:

$b(t) = 9^6 + 6^4t-6^3t^2$

We differentiate the given function.

Thus, the growth rate is given by:

$\displaystyle\frac{db(t)}{dt} = \frac{d}{dt}(9^6 + 6^4t-6^3t^2)\\\\= 6^4-2(6^3)t$

a) Growth rates at t = 0 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=0}= 6^4-2(6^3)(0) = 1296\text{ bacteria per hour}$

b) Growth rates at t = 3 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=3}= 6^4-2(6^3)(3) = 0\text{ bacteria per hour}$

c) Growth rates at t = 6 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=6}= 6^4-2(6^3)(6) = -1296\text{ bacteria per hour}$

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3 years ago