Find two numbers such that the sum of the first and three times the second is 5 and the sum of the second and two times the firs
t is 8.
1 answer:
Answer:
first number= 19/5
second number= 2/5
Step-by-step explanation:
first number= a
second number= b
a+3b=5 equation 1
b+2a=8 equation 2
using equation 2 we have
b=8-2a
using equation 1 we have
a+3(8-2a)=5
a+24-6a=5
24-5=6a-a
19=5a
a=19/5 = first number
so using b=8-2a
we have
b=8-2(19/5)
b=8-(38/5)
b=2/5 = second number
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3(4x+5)+2
3•4x=12x
5•3=15+2=17
Answer= 12x + 17
350 because of the concert I’m just kindling I need points
Rewriting the left hand side,
csc²t - cost sec t
= (1/sin²t)-(cost)(1/cost)
= 1/sin²t - 1
= 1/sin²t - sin²t/sin²t
= (1-sin²t)/sin²t
= cos²t/sin²t
= cot²t
Answer:
4x+12=120°
X=12-4=120°
X=8=120
X=120/8
X=15
That's the simplest lololol
