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2 years ago

How do we check if a solution is extraneous? Use these equations as examples to help you explain:

Mathematics

1 answer:

Maurinko [17]2 years ago

5 0

Multiplying both sides of a rational equation by a variable expression introduces the possibility of extraneous solutions. Therefore, we must check the solutions against the set of restrictions. If a solution is a restriction, then it is not part of the domain and is extraneous.

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Need help solving these, if anyone is able to help, then please feel free! Help please ❤

crimeas [40]

# 3 is .4 because u multiply .2*.2

4 0

3 years ago

How do I find Q15 c, d and e???

klasskru [66]

I would make a tree because there are not that many possibilities. to get 99 with numbers between 1 and 50 you have 50+49 and 49 and 50 nothing else will work because if one spin is less than 49 say 48 to get 99 you need 51 which is impossible. either you 50 then 49 or you get 49 then 50. the others are a bit harder because you need to find all possible spins that will sum to 52 which is a much larger number.

8 0

3 years ago

Solve 6 sin((π/5)x)=5 for the four smallest positive solutions

cupoosta [38]

Answer:

1.568, 3.432, 11.568, 13.432

Step-by-step explanation:

Divide equation $6\sin\left(\dfrac{\pi }{5}x\right)=5$ by 6:

$\sin\left(\dfrac{\pi }{5}x\right)=\dfrac{5}{6}.$

Then

$\dfrac{\pi }{5}x=(-1)^k\arcsin\left(\dfrac{5}{6}\right)+\pi k,\ k\in Z,$

$x=(-1)^k\dfrac{5}{\pi }\arcsin\left(\dfrac{5}{6}\right)+5k,\ k\in Z.$

Since $\arcsin\left(\dfrac{5}{6}\right)\approx 56^{\circ}\approx \dfrac{\pi }{3.2},$

four smallest positive solutions are

1. $\dfrac{5}{3.2}\approx 1.568,\ k=0;$

2. $\dfrac{-5}{3.2}+5\approx 3.432,\ k=1;$

3. $\dfrac{5}{3.2}+10\approx 11.568,\ k=2;$

4. $\dfrac{-5}{3.2}+15\approx 13.432,\ k=3.$

3 0

3 years ago

In preparation for the Maths festival you are asked to work out how much water the drinks area will need for the event. The aren

Tresset [83]

Total liters of water needed = number of attendee x water consumption/attendee wherenumber of attendee = 7000 attendeewater consumption/attendee = 0.5litre of water/attendeetotal = 7000 x 0.5 litre = 3500 litrescomputing the water requirement per 200 litre containerwater requirement in the event = 3500 litres/200 litre /container = 17.5 containersThe water requirement in the event at the arena of 7000 attendees is 17.5 containers.

3 0

3 years ago

Answer please need this quick

SOVA2 [1]

Answer: 6 units

Step-by-step explanation:

The line that goes through half of the circle at the top is the radius. Hope this helps!’

3 0

2 years ago