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Mathematics

2 answers:

anygoal [31]3 years ago

6 0

Answer:

Sample Response: First, the like terms had to be combined using the lowest common denominator (LCD). Then the subtraction property of equality was used to isolate the variable term. Finally, both sides of the equation were multiplied by the reciprocal of the coefficient to solve for a.

Step-by-step explanation:

r-ruslan [8.4K]3 years ago

3 0

Answer:

First, the like terms had to be combined using the lowest common denominator (LCD). Then the subtraction property of equality was used to isolate the variable term. Finally, both sides of the equation were multiplied by the reciprocal of the coefficient to solve for a.

Step-by-step explanation:

You might be interested in

If y=e5t is a solution to the differential equation

a_sh-v [17]

Answer:

k = 30, $y(t) = C_1e^{5t}+C_2e^{6t}$

Step-by-step explanation:

Since $y=e^{5t}$ is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that

$\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}$

Then, replacing the derivatives in the equation we have:

$25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0$

Since $e^{5t}$ is a positive function, we have that

$25-55+k = 0 \rightarrow k = 30$ .

Now, consider a general solution $y(t) = Ae^{rt}, A \in \mathbb{R}$ , then, by calculating the derivatives and replacing them in the equation, we get

$Ae^{rt}(r^2-11r+30)=0$

We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.

Therefore, the general solution is

$y(t) = C_1e^{5t}+C_2e^{6t}$

8 0

3 years ago

How to solve this problem.

andreyandreev [35.5K]

Notice the picture below

now, if the arc "x" is 48°, and the arc across the circle is also 48°, then those chords are congruent, notice the chords in red