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worty [1.4K]
2 years ago
13

Solve using the elimination method. 5r-2s=21 & 2r+5s=20

Mathematics
1 answer:
cestrela7 [59]2 years ago
5 0
I hope this is correct. I solved it using the addition elimination method. The answer is s= 2, and r= 5. I hope that was helpful.
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What is the number product of its prime factor 48
Vanyuwa [196]
The factor of 48 are : 1,2,3,4,6,8,12,16,24,48
The prime factor are : 2x2x2x2x3
5 0
3 years ago
5.37 + x =12.89 explanation please it’s two step equation
Wewaii [24]

Answer:

7.52

Step-by-step explanation:

5.37 + x = 12.89

In order to solve this equation, we must isolate the variable, x.

1.) To do this, subtract 5.37 on both sides of the equation.

5.37 + x = 12.89

-5.37         -5.37

x = 7.52

Therefore, x = 7.52

6 0
2 years ago
Read 2 more answers
A record store owner assesses customers entering the store as high school age, college age,
Eva8 [605]

Step-by-step explanation:

(a) P = (0.3)(0.2) + (0.5)(0.6) + (0.2)(0.8)

P = 0.52

(b) (0.5)(0.6) = 0.3

P = 0.3 / 0.52

P = 0.58

3 0
3 years ago
A stadium has 50,000 seats. Seats in section A cost $30, seats in section B cost $24, and seats in section C cost $18. the numbe
svetoff [14.1K]

Answer:

Section A = 25,000 seats

Section B = 14,600 seats

Section C = 10,400 seats

Step-by-step explanation:

Total Seats = 50,000

Seats in Section A cost = $30

Seats in Section B cost = $24

Seats in Section C cost = $18

Total sales from the event = $1,287,600

No. of Seats in section A = No. seats in Section B + No. seats in Section C

A = B + C

or, 2A = 50,000

A = 25,000 seats @ $30/seat = $750,000

B + C = 25,000

24B + 18C = 537,600

24B + 18(25,000 - B) = 537,600

24B + 450,000 - 18B = 537,600

6B = 87600

B = 14,600

C = 10,400

Hence;

A = 25,000 seats

B = 14,600 seats

C = 10,400 seats

7 0
3 years ago
Prove the trigonometric identity
Annette [7]

Answer:

Proved See below

Step-by-step explanation:

Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started

For this question we need to know the following identities:

1+tan^{2}x=sec^2x\\\\1+cot^2x=cosec^2x\\\\sin^2x+cos^2x=1

Lets solve the bottom most part first:

1-\frac{1}{1-sec^2x} \\\\

Take LCM

1-\frac{1}{1-sec^2x} \\\\\frac{1-sec^2x-1}{1-sec^2x} \\\\\frac{-sec^2x}{1-sec^2x} \\\\\frac{-(1+tan^2x)}{-tan^2x}

now break the LCM

\frac{-1}{-tan^2x}+\frac{-tan^2x}{-tan^2x}\\\\\frac{1}{tan^2x}+1\\\\cot^2x+1

because 1/tan = cot x

and furthermore,

cot^2x+1\\cosec^2x

now we solve the above part and replace the bottom most part that we solved with cosec^2x

\frac{1}{1-\frac{1}{cosec^2x} } \\\\\frac{1}{1-sin^2x} \\\\\frac{1}{cos^2x}\\\\sec^2x

Hence proved! :D

4 0
3 years ago
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