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Molodets [167]
4 years ago
15

In the diagram, AB = AC. Solve for x. Pls I need help!

Mathematics
1 answer:
Fiesta28 [93]4 years ago
4 0
ANSWER:

x = 11

OR

x = - 5

In triangle ABC displayed in the diagram, the value of ( x ) is 11 or - 5.

STEP-BY-STEP EXPLANATION:

In triangle ABC, AB is equivalent to AC. This means that the triangle is an isosceles triangle as two sides are equivalent in length.

Since triangle ABC is an isoceles triangle, the base angles must be equivalent to each other.

( x^2 - 6x )° and 55° are both base angles of the isosceles triangle ABC.

THEREFORE:

x^2 - 6x = 55

x^2 - 6x - 55 = 0

x^2 - 11x + 5x - 55 = 0

x ( x - 11 ) + 5 ( x - 11 ) = 0

( x - 11 ) ( x + 5 ) = 0

Solution No. 1 -

x - 11 = 0

x = 11

Solution No. 2 -

x + 5 = 0

x = - 5
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y(x)=2xe^{3x} (See attached graph)

Step-by-step explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation am^2+bm+c=0 where the values of m are the roots:

y''-6y'+9y=0\\\\m^2-6m+9=0\\\\(m-3)^2=0\\\\m-3=0\\\\m=3

Since the values of m are equal real roots, then the general solution is y(x)=C_1e^{m_1x}+C_2xe^{m_1x}.

Thus, the general solution for our given differential equation is y(x)=C_1e^{3x}+C_2xe^{3x}.

To account for both initial conditions, take the derivative of y(x), thus, y'(x)=3C_1e^{3x}+C_2e^{3x}+3C_2xe^{3x}

Now, we can create our system of equations given our initial conditions:

y(x)=C_1e^{3x}+C_2xe^{3x}\\ \\y(0)=C_1e^{3(0)}+\frac{C_2}{6}(0)e^{3(0)}=0\\ \\C_1=0

y'(x)=3C_1e^{3x}+C_2e^{3x}+3C_2xe^{3x}\\\\y'(0)=3C_1e^{3(0)}+C_2e^{3(0)}+3C_2(0)e^{3(0)}=2\\\\3C_1+C_2=2

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3C_1+C_2=2\\\\3(0)+C_2=2\\\\C_2=2

Thus, our final solution is:

y(x)=C_1e^{3x}+C_2xe^{3x}\\\\y(x)=2xe^{3x}

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