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3 years ago

What's the greatest common factor of 129 and 86

1 answer:

6 0

To get the Greatest Common Factor of 129 and 86 we need to factor each value first and then we choose all the copies of factors and multiply.

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In triangle ABC, m angle A = 30° and m angle B = 60°. Matches these in the order from smallest to largest. AB, BC, AC

trasher [3.6K]

Answer:

ab, ac, bc

Step-by-step explanation:

a=30, b=60, c=180-(30+60) = 180-90=90

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3 years ago

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What is 0.267 rounded off to 1 significant figure

Anuta_ua [19.1K]

The answer is 0.3

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3 years ago

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Complete the coordinate proof.

Lady_Fox [76]

Answer:

j RJ irjcirkcks he ed dew wh he x vhg

Step-by-step explanation:

jrjrirjcjrjifjrjfjjfwodoqldmfjrj

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3 years ago

How many lines of symmetry does this regular polygon have A.0 B.1 C.2 D.5

REY [17]

Answer:

Step-by-step explanation:

the top pair and the middle pair

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3 years ago

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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