MATHEMATICS HIGH SCHOOL

For an angle Θ with the point (5, -12) on its terminating side, what is the value of cosine

Answers

Answer 1
Answer: cosine = x-component/hypotenuse

cosine = 5 / √(12^2 + 5^2) = 5/13 = 0.385

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Melissa, of Melissa's Lawn and Landscaping Service, needs to replace her mowers.She has ordered four new gas mowers, at $499 each, and a negotiated Trade
Discount Rate of 9%, with terms of 2/10 EOM, and FOB Shipping Point. The
seller has agreed to prepay the shipping charge of $80, which is not included in the
invoice amount. The invoice from the manufacturer is dated July 26.
Determine:
A)
By what date does Melissa have to pay in order to be able to take the cash
discount?
B)
What is the payment amount assuming that Melissa pays before the end of
the discount period?

Answers

Answer:

A) 10th of August

B) $1780.03

Step-by-step explanation:

A) From the payment terms of 2/10 EOM, Melissa is to receive a 2% discount if she pays within 10 days from the end of the month

From the date on the invoice, July 26, the date at which Melissa has to pay to take the cash discount is before the 10th of August

B) The payment amount will be 2% off the original payment amount

The total cost of the four new gas mowers less the trade discount = 499 × 4 × (100 - 9)/100 =$1816.36

The 2% discount will then be $1816.36 × (100 - 2)/100 = $1780.03

The payment amount assuming Melissa pays before the end of the discount period = $1780.03.

55% of the 400 students at a school went to the game.Select true or false

A. 55 students went to the game.

B. 180 students did not go to the game.

C. More than half the students went to the game.

Answers

A. False

B. True

C. True

Of 60 students in a class 2/3 are girls, and 2/5 of the class are taking music lessons. What is the maximum number of girls that are not taking music lessons?

Answers

24 girls. Because 60/3=20 20*2=40. So than u do 40/5=8 8*2=16 40-16=24.

Using the laws of indices, simplify the following:-

√h × √h³ =

Answers

\sqrt h\cdot√(h^3)=\nh^{(1)/(2)}\cdot \left(h^3\right)^{(1)/(2)}=\nh^{(1)/(2)}\cdot h^{(3)/(2)}=\nh^{(1)/(2)+(3)/(2)}=\nh^{(4)/(2)}=\nh^2

or other way

\sqrt h\cdot√(h^3)=\n √(h^4)=\n √((h^2)^2)=\n |h^2|=\n h^2

Chromium-51 has a half-life of 28 days. If Georgia leaves a bottle of 10,000-milligram capsules in her medicine cabinet for a year, what will be the strength in milligrams of each capsule? Round to the nearest milligram

Answers

For finding the strength of the capsules after one year, we will use half-life formula. The formula is:

A = A₀ ((1)/(2))^(t)/(h)

where, A= Final amount

A₀ = Initial amount

t= time elapsed

h= half-life

Here, in this problem A₀ = 10000 milligram, t= 1 year or 365 days

and h= 28 days

So, A = 10000 ((1)/(2) )^(365)/(28)

⇒ A = 10000 ((1)/(2))^1^3^.^0^4

⇒ A = 1.187

So, the strength of the capsules after one year will be 1.187 milligrams.

(5+2 i)(4-3i) - (5-2yi)(4-3i) in a + bi form, where y is a real number.

Answers

(5+2 i)(4-3i) - (5-2yi)(4-3i)

Factorize out (4 -3i)

(4 -3i)( (5 +2i) - (5 -2yi) )

=  (4 -3i)(5 +2i - 5 + 2yi)

=  (4 -3i)(5 - 5 + 2i + 2yi)

= (4 -3i)(2i + 2yi)

= (4 - 3i)(2 + 2y)i.    Let's multiply the first two.

(4 - 3i)(2 + 2y) = 2*(4 -3i) + 2y*(4 - 3i)           
                      =  8 - 6i + 8y - 6yi
                      =  8 + 8y - 6i - 6yi

(4 - 3i)(2 + 2y)i = (8 + 8y - 6i - 6yi)i                Note: i² = -1
                       =  8i  + 8yi - 6i² - 6yi²
                       =  8i  +  8yi - 6(-1) - 6y(-1)  
                       =  8i  +  8yi  + 6 + 6y
                       =  6 + 6y + 8i + 8yi
                       =  (6 + 6y) + (8 + 8y)i    In the form a + bi
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