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Dmitriy789 [7]
3 years ago
6

The center of a circle lies on the line y=6x+2 and is tangent to the X axis at (-2,0) what is the equation of the circle in stan

dard form?
Mathematics
1 answer:
Murrr4er [49]3 years ago
8 0

Answer:

(x+2)^2 + (y+10)^2 = 100

Step-by-step explanation:

y = 6x + 2

Z(-2,0)

x = -2

y = 6 x (-2) + 2

y = -12 + 2

y = -10

M(-2, -10)

r = 0 - (-10) = 10

(x + 2)^2 + (y + 10)^2 = 100

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The bigger of two numbers is four more
Otrada [13]

Answer:

X=the smaller

y=the larger

The bigger of two numbers is four more than the smaller, then:

y=x+4

one more than twice the smaller is the same as the larger, then

1+2x=y

We have the following system of equations:

y=x+4

y=2x+1

We solve this system by equal values method:

x+4=2x+1

x-2x=1-4

-x=-3

x=3

We find the value of "y"now:

y=x+4

y=3+4

y=7

the values of x and y are:

x=3

y=7

Step-by-step explanation:

7 0
4 years ago
Select each pair of equivalent numbers. PLS HELP DUE TMR AND I DONT HOW TO SOLVE IT, PLS SHOW HOW U SET IT UP
BaLLatris [955]
The pairs that are equivalent to each other are 40/1000 and 40%, 6/5 and 120%, 1/8 and 12.5% . They are the correct answers because if we divide for example 6/5 , we would get 1.2 and to turn a decimal into a percent we have to move the decimal point 2 times to the right meaning that the percentage would be 120% and you just do that for all! I hope this helped :)!!
3 0
3 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

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

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
7/8x=42<br><br> What is the value of x?
Artist 52 [7]
\frac{7}{8}*x=42 \\ \\x=42: \frac{7}{8} \\ \\ x=42*  \frac{8}{7} \\ \\ x= \frac{336}{7} \\ \\ \boxed{x=48}
3 0
3 years ago
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