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1 year ago

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Determine which function has a greater initial value and explain how you know.

1 answer:

murzikaleks [220]1 year ago

8 0

Answer:

Function 2 has a greater initial valueStep-by-step explanation:An initial value of a function, aka y-intercept, is obtained by using x=0 and evaluating the function at that point. Evaluating the two functions given:Function1: Function2 has value 3 for x=0So, Function 2 has a greater initial valueFunction 2 has a greater initial valueStep-by-step explanation:An initial value of a function, aka y-intercept, is obtained by using x=0 and evaluating the function at that point. Evaluating the two functions given:Function1: Function2 has value 3 for x=0So, Function 2 has a greater initial value

Step-by-step explanation:

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Do the following side lengths form a right triangle? Explain. 18 ft, 23 ft, 29 ft

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Answer:

Explaination:

Step-by-step explanation:

B: yes; 18^2 + 29^2=1165(equal sign with slash) 23^2

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2 years ago

Mrs.Rodriguez is going to use 6 1/3 yards of material to make two dresses. The larger dress requires 3 2/3 yards of material. Ho

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2 years ago

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Put the numbers in order from least to greatest. - 29 , -5.39, - 27 5 , -5 7 20 A) - 27 5 , -5.39, - 29 , -5 7 20 B) - 29 , -5.3

lara [203]

Answer:

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2 years ago

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Prove the identity (n-5)^2 - (6n-35)=(n-10)(n-6)

elena-14-01-66 [18.8K]

Answer:

(n-5)^2 - (6n-35)=(n-10)(n-6)

-----------

n²-10n+25-6n+35 =
n²-16+60 = n²- 10n - 6n + 60 =
n(n-10) - 6(n-10) =
(n-10)(n-6)

7 0

3 years ago

Please someone help me...

laiz [17]

Step-by-step explanation:

First factor out the negative sign from the expression and reorder the terms

That's

$\frac{1}{ - (( \tan(2A) - \tan(6A) )} - \frac{1}{ \cot(6A) - \cot(2A) }$

<u>Using trigonometric </u><u>identities</u>

That's

<h3> $\cot(x) = \frac{1}{ \tan(x) }$ </h3>

<u>Rewrite the expression</u>

That's

$\frac{1}{ - (( \tan(2A) - \tan(6A) )} - \frac{1}{ \frac{1}{ \tan(6A) } } - \frac{1}{ \frac{1}{ \tan(2A) } }$

We have

<h3> $- \frac{1}{ \tan(2A) - \tan(6A) } - \frac{1}{ \frac{ \tan(2A) - \tan(6A) }{ \tan(6A) \tan(2A) } }$ </h3>

<u>Rewrite the second fraction</u>

That's

<h3> $- \frac{1}{ \tan(2A) - \tan(6A) } - \frac{ \tan(6A) \tan(2A) }{ \tan(2A) - \tan(6A) }$ </h3>

Since they have the same denominator we can write the fraction as

$- \frac{1 + \tan(6A) \tan(2A) }{ \tan(2A) - \tan(6A) }$

Using the identity

<h3> $\frac{x}{y} = \frac{1}{ \frac{y}{x} }$ </h3>

<u>Rewrite the expression</u>

We have

<h3> $- \frac{1}{ \frac{ \tan(2A) - \tan(6A) }{1 + \tan(6A) \tan(2A) } }$ </h3>

<u>Using the trigonometric identity</u>

<h3> $\frac{ \tan(x) - \tan(y) }{1 + \tan(x) \tan(y) } = \tan(x - y)$ </h3>

<u>Rewrite the expression</u>

That's

<h3> $- \frac{1}{ \tan(2A -6A) }$ </h3>

Which is

<h3> $- \frac{1}{ \tan( - 4A) }$ </h3>

<u>Using the trigonometric identity</u>

<h3> $\frac{1}{ \tan(x) } = \cot(x)$ </h3>

Rewrite the expression

That's

<h3> $- \cot( - 4A)$ </h3>

<u>Simplify the expression using symmetry of trigonometric functions</u>

That's

<h3> $- ( - \cot(4A) )$ </h3>

<u>Remove the parenthesis </u>

We have the final answer as

<h2> $\cot(4A)$ </h2>

As proven

Hope this helps you

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2 years ago

Read 2 more answers