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castortr0y [4]
3 years ago
5

How to solve for x and y

Mathematics
1 answer:
Minchanka [31]3 years ago
4 0

<u>Answer-</u>

<em>The values of </em><em>x and y are 20° and 30</em><em>, respectively.</em>

<u>Solution-</u>

As in the given triangle all the angles are same so it must a equilateral triangle.

In an equilateral triangle all the measurements of the angles and side length are same. The measurement of the angels are 60°.

As given in the question one side length is 46, so all the side length are same.

So,

\Rightarrow y+16=46\\\\\Rightarrow y=46-16\\\\\Rightarrow y=30

We know that an exterior angle of a triangle is equal to the sum of the opposite interior angles. 80° is the exterior angle opposite to x and one angle of triangle at the top vertex.

\Rightarrow 80^{\circ}=x+60^{\circ}

\Rightarrow x=80^{\circ}-60^{\circ}

\Rightarrow x=20^{\circ}

Therefore, the values of x and y are 20° and 30, respectively.

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The base angles theorem converse states if two angles in a triangle are congruent, then the sides opposite those angles are also congruent. The Isosceles Triangle Theorem states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle.

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I need help with number 30 and please show work
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10 x 6 = 60
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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
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