Question

100 POINTS+BRAINLIEST 8TH GRADE MATH PT 2: Word problem involving algebraic equations on them-As usual please explain the steps simply and some things to remember. answers for points will be reported :))

Answer 1

Answer:

$\sf a. \quad \dfrac{21}{10+x}+\dfrac{21}{10-x}=5$

Step-by-step explanation:

Formula for calculating speed:

$\boxed{\sf Speed=\dfrac{Distance}{Time}}$

Therefore:

$\boxed{\sf Time=\dfrac{Distance}{Speed}}$

Given values:

• x = the current (in mph)
• Speed = 10 mph
• Distance = 21 miles
• Total time = 5 hours

If the boat is traveling downstream, the current (x mph) will be pushing the boat faster, and so the boat's speed will increase by x mph.

Therefore, the expression for the time it takes for the boat to travel downstream is:

$\sf Time\:(downstream)=\dfrac{21}{10+x}$

If the boat is traveling upstream, the current (x mph) will be pushing against the boat and so the boat's speed will decrease by x mph.

Therefore, the expression for the time it takes for the boat to travel upstream is:

$\sf Time\:(upstream)=\dfrac{21}{10-x}$

Therefore, if the total time it takes to travel downstream and upstream is 5 hours:

$\implies \sf \dfrac{21}{10+x}+\dfrac{21}{10-x}=5$

To solve for x:

$\implies \sf \dfrac{21}{10+x}+\dfrac{21}{10-x}=5$

$\implies \sf \dfrac{21(10-x)}{(10+x)(10-x)}+\dfrac{21(10+x)}{(10-x)(10+x)}=5$

$\implies \sf \dfrac{21(10-x)+21(10+x)}{(10+x)(10-x)}=5$

$\implies \sf21(10-x)+21(10+x)=5(10+x)(10-x)$

$\implies \sf 210-21x+210+21x=5(100-10x+10x-x^2)$

$\implies \sf 420=500-5x^2$

$\implies \sf 5x^2=80$

$\implies \sf x^2=16$

$\implies \sf x=\pm4$

Therefore, the rate of the current is 4 mph.

Answer 2

The answer is A for this