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nalin [4]
3 years ago
13

I need help with my ixl ..

Mathematics
1 answer:
Ad libitum [116K]3 years ago
6 0

Answer:

The measure of angle E is 45 degrees

Step-by-step explanation:

To get the measure if angle E, we use the appropriate trigonometric ratio

From the question, EF is the hypotenuse since it faces the right angle

EG is adjacent to angle E

so we are going to use the trigonometric ratio that connects adjacent to the hypotenuse

This trigonometric ratio is the cosine

It is the ratio of the adjacent to the hypotenuse

Thus,

we have it that;

Cos E = EG/EF

cos E = 6√26/12√13

E = arc Cos ( 6 √26/12 √13)

E = 45 degrees

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Answer: PSR (Perfect Square Rule) = (x-3/2)^2

The you simplify.... = x^2 - 3x + 9/4

Step-by-step explanation: Factor by using the PSR (Perfect Square Rule). Then you have to simplify.

Hope this helps you out! ☺

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3 years ago
Explain how to solve the problem below. In your response, you must analyze the given information, discuss a strategy or plan to
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Answer:

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2 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
Explain why f(x) = x^2+4x+3/x^2-x-2 is not continuous at x = -1.
liberstina [14]

Answer:

The value of x = -1 makes the denominator of the function equal to zero. That is why this value is not included in the domain of f(x)

Step-by-step explanation:

We have the following expression

f(x) = \frac{x^2+4x+3}{x^2-x-2}

Since the division between zero is not defined then the function f(x) can not include the values of x that make the denominator of the function zero.

Now we search that values of x make 0 the denominator factoring the polynomial x^2-x-2

We need two numbers that when adding them get as a result -1 and when multiplying those numbers, obtain -2 as a result.

These numbers are -2 and 1

Then the factors are:

(x-2) (x + 1)

We do the same with the numerator

x^2+4x+3

We need two numbers that when adding them get as a result 4 and when multiplying those numbers, obtain 3 as a result.

These numbers are 3 and 1

Then the factors are:

(x+3)(x + 1)

Therefore

f(x) = \frac{(x+3)(x+1)}{(x-2)(x+1)}

Note that \frac{(x+1)}{(x+1)}=1 only if x \neq -1

So since x = -1 is not included in the domain the function has a discontinuity in x = -1

3 0
3 years ago
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