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3 years ago

Write the slope-intercept form of the equation of the line given the slope and the y-intercept.

Mathematics

1 answer:

Readme [11.4K]3 years ago

8 0

Answer:

The answer is y=-8/3x + 4

Step-by-step explanation:

There is this format called linear function where it looks like this

y=mx + b

where m is the slope and b is the y-intercept

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The question is below the screen

sesenic [268]

9514 1404 393

Answer:

x = 12

Step-by-step explanation:

The square of the length of the tangent segment is equal to the product of lengths from the external common point to the two circle intersections of the secant.

x^2 = (2)(70+2)

x^2 = 144

x = √144

x = 12

4 0

3 years ago

Can u get the steps to solving <br> -2(4)^2+4

svp [43]

(4)^2 must be done first.

That’ll give you 16. (4x4)

16 x -2= -32

-32+4= -28

8 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

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