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3 years ago

What happens to the Law of Cosines if the known angle is 90 degrees?

Mathematics

1 answer:

poizon [28]3 years ago

8 0

Answer:

See below.

Step-by-step explanation:

It transforms to the Pythagoras theorem.

c^2 = a^2 + b^2 - 2ab cos C

If C = 90, cos C = zero so the last term disappears.

c^2 = a^2 + b^2 - 2ab * 0

c^2 = a^2 + b^2.

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Brody makes $38,000 per year no matter how many hours he works. Shonda makes $18 per hour and works 40 hours per week. Which one

Ann [662]

Answer:

64.00 it is 64.00

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

An airplane has a maximum capacity of 118 passengers. The flight attendant has loaded 40 passengers. Which inequality represents

kupik [55]

Answer:

118-40=78

p either is less tham or equal to 78

the answer is c

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2 years ago

Eric was rock climbing. At one point, he stopped and climbed straight down 2\dfrac{1}{2}2 2 1 2, start fraction, 1, divided by

podryga [215]

Answer:

Hence, Eric's elevation changed $4\frac{1}{4}$ meters above. So after the 2 moves, he is up by $4\frac{1}{4}$ meters.

Step-by-step explanation:

Let us assume "negative" as climbing up and "positive" as climbing up. Here we need to find the change in elevation.

$2\frac{1}{2}$ meters down means $-2\frac{1}{2}$ .
$6\frac{3}{4}$ meters up means $+6\frac{3}{4}$ .

Now we have:

$=-2\frac{1}{2} + 6\frac{3}{4}\\\\=-\frac{5}{2} +\frac{27}{4} \\\\=\frac{-5(2)+27}{4} \\\\=\frac{-10+27}{4} \\\\=\frac{17}{4} = 4\frac{1}{4}$

4 0

2 years ago

Arleen has a gift card for a local lawn and garden store. She uses the gift card to rent a tiller for 3 days. It costs $48 per d

MrRa [10]

Given :

Arleen has a gift card for a local lawn and garden store. She uses the gift card to . It costs $48 per day to rent the tiller. She also buys a rake for $5.

To Find :

The value on her gift card decreased by how many dollars.

Solution :

Decrease in dollar is given by :

$D =\text{ rent a tiller for 3 days+ price of rake}\\\\D= \$( 48\times 3 + 5)\\\\D = \$149$

Therefore, the value on her gift card decreased by 149 dollars.

Hence, this is the required solution.

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3 years ago

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 i

stich3 [128]

Complete question :

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 inches. Round answers to the nearest tenth of a percent. a) What percentage of years will have an annual rainfall of less than 43 inches? b) What percentage of years will have an annual rainfall of more than 39 inches? c) What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

Answer:

0.61053

0.66314

0.26647

Step-by-step explanation:

Given that :

Mean (m) = 41.4 inches

Standard deviation (s) = 5.7 inches

a) What percentage of years will have an annual rainfall of less than 43 inches?

P(x < 43)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (43 - 41.4) / 5.7 = 0.2807017

p(Z < 0.2807) = 0.61053 ( Z probability calculator)

b) What percentage of years will have an annual rainfall of more than 39 inches? c)

P(x > 39)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (39 - 41.4) / 5.7 = −0.421052

p(Z > −0.421052) = 0.66314 ( Z probability calculator)

What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

P(x < 38)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (38 - 41.4) / 5.7 = −0.596491

p(Z < −0.596491) = 0.27545 ( Z probability calculator)

P(x < 42)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (42 - 41.4) / 5.7 = 0.1052631

p(Z < 0.1052631) = 0.54192 ( Z probability calculator)

0.54192 - 0.27545 = 0.26647

6 0

2 years ago