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4 years ago

5

the average life span of a lion is one year more than twice the life span of a fox , which is seven years Let i represent the li

ge span of a lio.

1 answer:

Snezhnost [94]4 years ago

8 0

The lift spam is 15 years if you are looking for an equation its is 2f+1=i

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A naples daily news feature article emphasized the number of runners who were "20-something". what percentage of the runners wer

Artemon [7]

Answer:

however the meaning of math

Step-by-step explanation:

20+20

7 0

3 years ago

Estimate the unit rate of 12 pairs of socks sell for $5.79 any body knows what the answer is?

Minchanka [31]

Answer:

About $0.24 per sock

Step-by-step explanation:

12 pairs = 24 socks

5.79/24 = 0.24125

8 0

4 years ago

Read 2 more answers

Find sin(2x), cos(2x) and tan (2x) from the given information. cot(x)=2/3, x in quadrant I

Usimov [2.4K]

Answer:

1.) 6√13/13

2.) 4√13/13

3.) 3

Step-by-step explanation:

given that cot(x) = 2/3

But cot(x) = 1/tan(x)

Substitutes 1/tan(x) for cot(x)

1/tan(x) = 2/3

Reciprocate both sides

Tan(x) = 3/2 = opposite/adjacent

Use pythagorean theorem to find the hypothenus.

Hypothenus = sqrt ( 3^2 + 2^2 )

Hypothenus = sqrt ( 9 + 4 )

Hypothenus = sqrt (13)

Hypothenus = √13

1.) Sin(x) = opposite/hypothenus = 3/√13

Rationalise

3/√13 × √13/√13

3√13/13

Sin(2x) = 2 × 3√13/13

Sin(2x) = 6√13/13

2.) Cos( x ) = adjacent/hypothenus

Cos (x) = 2/√13

Rationalise

2/√13 × √13 /√13

2√13/13

Cos(2x) = 2 × 2√13/13

Cos(2x) = 4√13/13

3.) Tan (x) = 3/2

tan (2x) = 2 × 3/2

Tan(2x) = 3

4 0

3 years ago

Please help (BRAINLYEST)There are 18 boys in a party. Of these,8 played darts, 4 jumped on a trampoline, and 16 did both. Consid

Anettt [7]

1) 8/9. This is a probability question.

4 0

3 years ago

Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))

e-lub [12.9K]

$sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0$

We can replace sin x with x anywhere in the limit as long as x approaches 0.

Also,

$\large \lim_{ x \to 0 } ~ x^x = 1$

I will make the assumption that <span>log(x)=ln(x)</span><span>.

The limit result can be proven if the base of </span><span>log(x)</span><span> is 10.
</span>
$\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}$

We get the indeterminate form 0/0, so we have to use <span>Lhopitals rule

</span> $\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1$
<span>
Therefore,

</span> $\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } =\boxed{ -1}$ <span>
</span>

3 0

3 years ago