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3 years ago

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classify the segment in each part below. only use the information given in the figure do not assume measures are equal unless th

e markings indicate they are

1 answer:

tensa zangetsu [6.8K]3 years ago

6 0

Answer:

i dont know i just need points for more questions sorry

Step-by-step explanation:

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6. RVHS sold 143 Adult and Student tickets to the

slava [35]

Answer:

57 adult tickets, 86 student tickets

7 0

3 years ago

Given that<br> x:8=2:x<br> Calculate the positive value of x.

kow [346]

Answer:

$x = 4$

Step-by-step explanation:

$x : 8 = 2 : x \\ \frac{x}{8} = \frac{2}{x} \\ \\ cross \: \: multiple \\ {x}^{2} = 16 \\ x = \sqrt{16} \\ x = 4$

4 0

3 years ago

Read 2 more answers

What is the intercept form of the equation y=6x^2+12x-144

Stells [14]

Answer:

y = 6(x + 6)(x - 4).

Step-by-step explanation:

y = 6x^2 + 12x - 144

y = 6(x^2 + 2x - 24) Factoring the trinomial in the parentheses ,we get:

y = 6(x + 6)(x - 4)

4 0

3 years ago

(-8)x...x(-8) times 20

Stella [2.4K]

Answer:

1280

Step-by-step explanation:

rewrite (-8) times (-8) as an equivalent addition problem which is 8 times 8 = 64

take 64 then times it by 20 which is 64 times 20 equals 1,280

Hope this helps:)

6 0

3 years ago

Can you please answer the question?

Roman55 [17]

The <em>trigonometric</em> expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the <em>trigonometric</em> expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that <em>one</em> side of the equality is equal to the expression of the <em>other</em> side, requiring the use of <em>algebraic</em> and <em>trigonometric</em> properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}$

$\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}$

$\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}$

$\tan \alpha + 1 + \cot \alpha$

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1$

$\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1$

$\frac{1}{\cos \alpha \cdot \sin \alpha} + 1$

$\sec \alpha \cdot \csc \alpha + 1$

The <em>trigonometric</em> expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the <em>trigonometric</em> expression $\sec \alpha \cdot \csc \alpha + 1$ .

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

6 0

2 years ago