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3 years ago

How do you find the Y intercept ? I’m having some trouble

Mathematics

2 answers:

NISA [10]3 years ago

7 0

Answer:

so on a graph you have two axis's: the x (horizontal) and y (vertical). The y intercept is the point at which your line touches the y axis which is also when x=0. So your y intercept will always be (0,y) in this case when x is t and when t is 0, C (or y) is 8. x is your input (independent) and y is your output (dependent). The y intercept in the above model can be written as (0,8) or m=8

Step-by-step explanation:

marishachu [46]3 years ago

4 0

Answer:

The y intercept is always (0,y) so in this case it is (0,8)

Step-by-step explanation:

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Can someone help me with this?

Anna35 [415]

Answer:

Let 'a' be the first term, 'r' be the common ratio and 'n' be the number of terms

Series = 2+6+18.......= 2+2•3¹+ 2•3².......= 728

Now,

$Sum = \frac{a( {r}^{n} - 1) }{(r - 1)} \\$

So,

$\frac{a( {r}^{n} - 1)}{(r - 1)} = 728 \\ \frac{2( {3}^{n} - 1) }{(3 - 1)} = 728 \\ \frac{2( {3}^{n} - 1) }{2} = 728 \\ {3}^{n} - 1 = 728 \\ {3}^{n}=728+1\\ {3}^{n} = 729 \\ {3}^{n} = {3}^{6} \\ \boxed{ n = 6}$

Therefore, number of terms is 6

6 is the right answer.

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

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