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Leno4ka [110]
2 years ago
12

A triangle ABC is inscribed in a circle, such that AB is a diameter. What are the measures of angles of this triangle if: measur

e of arc BC = 134°;
Mathematics
1 answer:
shutvik [7]2 years ago
4 0

Answer:

The measures of angles of this Δ are 23° , 67° , 90°

Step-by-step explanation:

* Lets talk about some facts in the circle

- An inscribed angle is an angle made from points sitting on the

 circle's circumference

- A central angle is the angle formed when the vertex is at the center

 of the circle

- The measure of an arc of a circle is equal to the measure of the

 central angle that intercepts the arc.

- The measure of an inscribed angle is equal to 1/2 the measure of

 its intercepted arc

- An angle inscribed across a circle's diameter is always a right angle

- The triangle is inscribed in a circle if their vertices lie on the

  circumference of the circle, and their angles will be inscribed

  angles in the circle

* Now lets solve the problem

- Δ ABC is inscribed in a circle

∵ its side AB is a diameter of the circle

∵ Its vertex C is on the circle

∴ ∠C is inscribed and across the circle's diameter

∴ ∠C is a right angle

∴ m∠C = 90°

∵ The measure of arc BC = 134°

∵ ∠A is inscribed angle subtended by arc BC

∵ The measure of an inscribed angle is equal to 1/2 the measure

   of its intercepted arc

∴ m∠A = 1/2 × 134° = 67°

∵ The sum of the measures of the interior angles of a triangle is 180°

∵ m∠A = 67°

∵ m∠C = 67°

∵ m∠A + m∠B + m∠C = 180°

∴ 67° + m∠B + 90° = 180°

∴ 157° + m∠B = 180° ⇒ subtract 157 from both sides

∴ m∠B = 23°

* The measures of angles of this Δ are 23° , 67° , 90°

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Rectangle ABCD with coordinates A(1,1), B(4,1), C(4,2) and D(1,2) dilates with respect to the origin to give rectangle A’B’C’D’.
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Answer: Option C.

Step-by-step explanation:

You need to find the distance AB with the formula for calculate the distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Then, substituting the coordinates of the points A(1,1) and B(4,1), you get:

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Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f(x) is a real n
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Answer:

See Below

Step-by-step explanation:

The function is a piecewise function defined as:

N(t)=\left \{ {{25t+150} \ 0 \leq t \leq 6 \atop {\frac{200+80t}{2+0.05t} \ t \geq 8}} \right.

a)

We need to find the limit of the function as t goes to infinity. This means what is the max value of fish in the pond given times goes to infinity (on an on).

We will take the 2nd part of the equation since t falls into that range, t is infinity, which is definitely greater than 8.

\lim_{t \to \infty} \frac{200+80t}{2+0.05t} \\\lim_{n \to \infty} \frac{80t}{0.05t}\\ \lim_{n \to \infty} \frac{80}{0.05} =1600

This means the maximum number of fish at this pond is 1600, no matter how long it goes on.

b)

A function is continuous at a point if we have the limit and the functional value at that point same.

Functional value at t = 8 is (we use 2nd part of equation):

\frac{200+80t}{2+0.05t}\\\frac{200+80(8)}{2+0.05(8)}\\=350

We do have a value and limit also goes to this as t approaches 8.

So, function is continuous at t = 8

c)

We want to find is there a "time" when the number of fishes in the pond is 250, during t from 0 to 6. We plug in 250 into N(t) and try to find t. Make sure to use the 1st part of the piece-wise function. Shown below:

N(t)=25t+150\\250=25t+150\\25t=250-150\\25t=100\\t=4

The time is 4 years when the number of fishes in the pond is 250

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