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2 years ago

I need help with finding the arc measure of this problem.

Mathematics

1 answer:

Leto [7]2 years ago

4 0

Answer:

180+69=249

I got 180 from the ABC, that is half the circle then + the 69

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What is question 2 and 3?

WARRIOR [948]

Number 2 is 1.00 and Number 3 is 1.50

4 0

3 years ago

Read 2 more answers

In an obtuse isosceles triangle the angle measurements are, x∘, x∘, and (10x−2)=128∘. Find the measurement of one of the acute a

klemol [59]

Answer:

26°

Step-by-step explanation:

An obtuse triangle is a triangle that has one obtuse angle. Obtuse angle is an angle that is greater than 90 degrees but less than 180 degrees.

An isosceles triangle is a triangle that has two equal sides and angles.

Therefore, an obtuse isosceles triangle is a triangle with an obtuse angle and two equal sides that have two equal acute angles (angle less than 90° ).

Given:

The three angles of the triangle are given to be x°, x° and (10x−2) = 128°. The obtuse angle is 128°, the two x° are acute angles. We are not using equation 10x − 2 since the value of the obtuse angle has been given as 128°

The sum of angles in a triangle is 180°

∴ x° + x° + 128° = 180°

2x° = 180° - 128°

2x° = 52°

x° = 52° / 2

x° = 26°

The measurement of one of the acute angles is 26°

7 0

2 years ago

Which portion of the unit circle satisfies the trigonometric inequality cos^2theta + sin^2theta is greater than or equal to 1. A

liberstina [14]

Answer:

Only points on the circle satisfy the given inequality.

Step-by-step explanation:

Given: Unit circle

To find: portion of the unit circle which satisfies the trigonometric inequality $\sin ^2\theta +\cos ^2\theta \geq 1$

Solution:

In the given figure, OA = 1 unit (as radius of the unit circle equal to 1)

$\sin \theta$ = side opposite to $\theta$ /hypotenuse

$\cos \theta$ = side adjacent to $\theta$ /hypotenuse

$\sin \theta =\frac{AB}{AO}\\\sin \theta =\frac{AB}{1}\\AB=\sin \theta$

$\cos \theta=\frac{OB}{AO}\\\cos \theta =\frac{OB}{1}\\OB=\cos \theta$

So, coordinates of A = $\left ( \cos \theta ,\sin \theta \right )$

For any point (x,y) on the unit circle with centre at origin, equation of circle is given by $x^2+y^2=1$

Put $(x,y)=\left ( \cos \theta ,\sin \theta \right )$

$\cos ^2\theta +\sin ^2\theta =1$

So, $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ satisfies the equation $x^2+y^2=1$

For points $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ inside the circle, $\cos ^2\theta +\sin ^2\theta$

For points $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ outside the circle, $\cos ^2\theta +\sin ^2\theta >1$

So, only points on the circle satisfy the given inequality.

4 0

2 years ago

Read 2 more answers

Y = −5x + 3<br><br> slope = −5<br> −3<br> 5

Aleks04 [339]

First ypu solve the yintercept amd then the x intercept

8 0

3 years ago

find continued product of (x+y) (x-y) (<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}"

leonid [27]

Answer:

$\large\boxed{(x+y)(x-y)(x^2+y^2)(x^4+y^4)=x^8-y^8}$

Step-by-step explanation:

$(x+y)(x-y)(x^2+y^2)(x^4+y^4)\\\\\text{use}\ (a-b)(a+b)=a^2-b^2\qquad(*)\\\\=\underbrace{(x+y)(x-y)}_{(*)}(x^2+y^2)(x^4+y^4)\\\\=\underbrace{(x^2-y^2)(x^2+y^2)}_{(*)}(x^4+y^4)\\\\=\left((x^2)^2-(y^2)^2\right)(x^4+y^4)\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\underbrace{(x^4-y^4)(x^4+y^4)}_{(*)}\\\\=\left(x^4\right)^2-\left(y^4\right)^2=x^8-y^8$

7 0

3 years ago