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3 years ago

If T is the midpoint of SU,find x. ( ST=8x+11 and TU= 12x-1)

Mathematics

2 answers:

sukhopar [10]3 years ago

8 0

Answer:

Step-by-step explanation:

Set the equations equal to each other and solve for x

8x+11 = 12x-1

11=4x-1

12=4x

x=3

Y_Kistochka [10]3 years ago

3 0

<h3>Answer: x = 3</h3>

============================================

Explanation:

T is the midpoint of SU, so point T cuts segment SU into two equal halves. Those two halves being ST and TU, therefore, ST = TU

ST = TU

8x+11 = 12x-1 ... substitution

8x+11-8x = 12x-1-8x ... subtract 8x from both sides

11 = 4x-1

11+1 = 4x-1+1 .... add 1 to both sides

12 = 4x

4x = 12

4x/4 = 12/4 .... divide both sides by 4

x = 3

As a check, plug x = 3 into each equation below

ST = 8x+11 = 8*3+11 = 24+11 = 35

TU = 12x-1 = 12*3-1 = 36-1 = 35

Each piece ST and TU is 35 units long. So this confirms we have the right answer.

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Step-by-step explanation: We are given to solve the following differential equation :

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Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

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