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3 years ago

Pls help me with this true or false question

Mathematics

2 answers:

dangina [55]3 years ago

7 0

Answer:

from my understanding true

Step-by-step explanation:

AlexFokin [52]3 years ago

7 0

Answer:

I think it is true

Step-by-step explanation:

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QUESTION IS DOW BELOW 5 POINTS EACH PLEASE HELP PLEASE HELP PLEASE HELP

masha68 [24]

The central angle in the circle is ∠DAC,major arc is BED, minor arc is ADC and BC=(5π*BD)/18.

Given that BD is diameter of the circle and angle BAC is 100°.

We are required to find the central angle, major arc, minor arc, m BEC, BC.

Angle is basically finding out the intensity of inclination of something on the surface.

In the circle central angles are many like BAC and CAD. We can write CAD as DAC also.

Major arc of a circle is that arc whose length is larger than all other arcs in the circle.

In our circle the major arc is arc BED.

Minor arc of a circle is that arc whose length is smaller.

In our circle the minor arc is arc ADC.

We know that arc's length is 2πr(Θ/360)

In this way BC=2π*(BD/2)*100/360

=(5π*BD)/18

We cannot find angle BEC.

Hence the central angle in the circle is ∠DAC,major arc is BED, minor arc is ADC and BC=(5π*BA)/18.

Learn more about angles at brainly.com/question/25716982

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2 years ago

Terrence McGraw sells lumber. He earns 4% commission on the first $5,000 in sales, 8% on the next $5,000, and 12% on sales over

Sindrei [870]

His total commission was $1560.

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3 years ago

Find the missing term of the following sequence. . . . –45, __, –12, . . .

julsineya [31]

<span>(-45 + (-12) / 2 = -28.5
</span>So the answer is -28.5

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3 years ago

For the right triangle shown, the lengths of two sides are given. find the third side. leave your answer in simplified, radical

lisov135 [29]

If c is the hypotenuse and b is the leg, then the third side (a):
$a= \sqrt{c^2-b^2}= \sqrt{16^2-9^2}= \sqrt{256-81}= \sqrt{175}=5 \sqrt{7}$

If b and с are legs, then hypotenuse (a):
$a= \sqrt{b^2+c^2}= \sqrt{9^2+16^2}= \sqrt{81+256}= \sqrt{337}$

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3 years ago

Read 2 more answers

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago