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aniked [119]
3 years ago
11

The triangle below is equilateral. Find the length of side 2 in simplest radical form

Mathematics
1 answer:
Masteriza [31]3 years ago
5 0

Answer:

x=\frac{4\sqrt{3}}{3}

Step-by-step explanation:

From the figure attached,

ΔABC is an equilateral triangle.

Therefore, by the property of the equilateral triangle,

Measure of each angle = 60°

By applying tangent rule in the right triangle ADC,

\text{tan}(60^{\circ})=\frac{\text{Opposite side}}{\text{Adjacent side}}

\sqrt{3}=\frac{4}{x}

x=\frac{4}{\sqrt{3}}

x=\frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3} }

x=\frac{4\sqrt{3}}{3}

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The product of a scalar and a matrix is found by multiplying each element of the matrix by the scalar. Multiply each element by -4.

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3 years ago
Ted and Fred are the owners of a gas station. They invested $150,000 each and pay an employee named Lawrence $35,000 per year. T
Serjik [45]

Answer:

Ted and Fred

Step-by-step explanation:

As per the question,

Ted and Fred are the owners of a gas station that is they are the owner of their own firm.

As being the owner of their own company, they are the one who is responsible for their loss. That is the firm’s owners always suffer the firm’s risk.

Business Risk: Risk that a business will not be able to cover its operating costs.

So in this case, Lawrence is the employee and legally he is not responsible to suffer any kind of the loss.

So he must be paid his salary  on time and in full whether or not the firm is running a profit, a loss, or just breaking  even.

As being owner Ted and Fred are responsible to withstand the business risk.

Hence, the person who is legally responsible for bearing the $40,000 loss is Ted and Fred.

8 0
3 years ago
What are the coordinates of the inflection point on the graph of y=(x+1)arctanx
Dennis_Churaev [7]

Inflection point is the point where the second derivative of a graph is zero.

y = (x+1)arctan xy' = (x+1)(arctan x)' + (1)arctan xy' = (x+1)/(x^2+1) + arctan xy'' = (x+1)(1/(1+x^2))' + 1/(1+x^2) + 1/(1+x^2)y'' = (x+1)(-1/(1+x^2)^2)(2x)+2/(1+x^2)y'' = ((x+1)(-2x)+1+x^2)/(1+x^2)^2y'' = (-2x^2-2x+2+2x^2)/(1+x^2)^2y'' = (-2x+2)/(1+x^2)^2

Solving for point of inflection: y'' = 00 = (-2x+2)/(1+x^2)^20 = -2x+2x = 1y(1) = (1+1)arctan(1) = 2 * pi/4 = pi/2

Therefore, E(1, pi/2).


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0
3 years ago
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