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3 years ago

How do i simplify question c?

Mathematics

2 answers:

Troyanec [42]3 years ago

7 0

Hope this will satisfy you

Alecsey [184]3 years ago

4 0

Combine the rational expressions using the least common denominator. =
2x/x+2

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If cos(x)=1/4 what is sin(x) and tan(x)

anygoal [31]

You can use the identity
cos(x)² +sin(x)² = 1
to find sin(x) from cos(x) or vice versa.

(1/4)² +sin(x)² = 1
sin(x)² = 1 - 1/16
sin(x) = ±(√15)/4

Then the tangent can be computed as the ratio of sine to cosine.
tan(x) = sin(x)/cos(x) = (±(√15)/4)/(1/4)
tan(x) = ±√15

There are two possible answers.
In the first quadrant:
sin(x) = (√15)/4
tan(x) = √15

In the fourth quadrant:
sin(x) = -(√15)/4
tan(x) = -√15

3 0

3 years ago

Some one help please algebra ..<br>-6=3x-6y<br>4x=4+5x

yanalaym [24]

X=2y-2

x=-4

hope i helped

5 0

3 years ago

Read 2 more answers

Distribute and Simplify: (p + 3)5

Tanzania [10]

Answer:

5p+5

Explanation:

5(p+3)

(5*p)+(5*3)

5p+5

4 0

3 years ago

A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe

adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

$\frac{dC}{dt}=Ci*Qi-Co*Qo$

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

$\frac{dC}{dt}=-Co*Qo$

Rearranging the equation, it becomes

$\frac{dC}{C}=-Qo*dt$

Integrating both sides

$\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}$

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

$C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\$

The final equation for the concentration of salt at any given time is

$C=exp^{-3*t-0.693}$

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

$C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}= -\frac{-0.266}{3}=0.088$

5 0

3 years ago

At noon, ship a is 30 nautical miles due west of ship

Dmitry_Shevchenko [17]

Let the origin of the coordinate system be the position of ship B at noon. Then the distance between the ships as a function of t (in hours) is
.. d = √((30 +23t)² +(17t)²)
.. = √(818t² +1380t +900)

The rate of change of distance with respect to time is
.. d' = (1/2)(2*818*t +1380)/√(818t² +1380t +900)
.. = (818t +690)/√(818t² +1380t +900)

At t=4, this is
.. d' = (818*4 +690)/√(818*16 +1380*4 +900)
.. = 3962/√19508
.. ≈ 28.37 . . . . . knots

The distance between the ships is increasing at about 28.37 knots at 4 pm.

8 0

3 years ago