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2 years ago

Can you please help me understand how to solve a trig ratio like this?

Mathematics

1 answer:

yKpoI14uk [10]2 years ago

6 0

Answer:

cos47° = $\frac{x}{7}$

Step-by-step explanation:

cos47° = $\frac{adjacent}{hypotenuse}$ = $\frac{x}{7}$

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A particular geometric sequence has strictly decreasing terms. After the first term, each successive term is calculated by multi

Maksim231197 [3]

Answer:

6 possible integers

Step-by-step explanation:

Given

A decreasing geometric sequence

$Ratio = \frac{m}{7}$

Required

Determine the possible integer values of m

Assume the first term of the sequence to be positive integer x;

The next sequence will be $x * \frac{m}{7}$

The next will be; $x * (\frac{m}{7})^2$

The nth term will be $x * (\frac{m}{7})^{n-1}$

For each of the successive terms to be less than the previous term;

then $\frac{m}{7}$ must be a proper fraction;

This implies that:

$0 < m < 7$

Where 7 is the denominator

The sets of $0 < m < 7$ is $\{1,2,3,4,5,6\}$ and their are 6 items in this set

Hence, there are 6 possible integer

3 0

2 years ago

34/23 as a decimal rounded to the nearest tenth

Ainat [17]

Answer:

1.5

Step-by-step explanation:

1.47826086

rounded to the nearest tenth is 1.5

4 0

2 years ago

A dieter believes that the average number of calories in a homemade peanut better cookie is more than the average number of calo

zalisa [80]

The confidence interval formula is computed by:

Xbar ± Z s/ sqrt (n)

Where:

Xbar is the mean

Z is the z value

S is the standard deviation

N is the number of samples

So our given are:

90% confidence interval with a z value of 1.645

Sample size 40, 45

Mean 180, 179

Standard deviation 2, 4

So plugging that information in the data will give us a confidence interval:

For 1:

Xbar ± Z s/ sqrt (n)

= 180 ± 1.645 (2 / sqrt (40))

= 180 ± 1.645 (0.316227766)

= 180 ± 0.520194675

= 179.48, 180.52

For 2:

Xbar ± Z s/ sqrt (n)

= 179 ± 1.645 (4 / sqrt (45))

= 179 ± 1.645 (0.596284794)

therefore, the answer is letter b

8 0

3 years ago

Evaluate lim x→∞ (3x+1)^(4/x), using l'hospital's rule as needed. show all work using proper notation. as you show your work, if

Alborosie

$\displaystyle\lim_{x\to\inty}(3x+1)^{4/x}=\lim_{x\to\infty}e^{\ln(3x+1)^{4/x}}=e^{\lim\limits_{x\to\infty}\ln(3x+1)^{4/x}}$

$\displaystyle\lim_{x\to\infty}\ln(3x+1)^{4/x}=\lim_{x\to\infty}\frac{4\ln(3x+1)}x\stackrel{\mathrm{LHR}}=\lim_{x\to\infty}\frac{4\frac3{3x+1}}1=\lim_{x\to\infty}\frac{12}{3x+1}=0$

$\implies\displaystyle\lim_{x\to\infty}(3x+1)^{4/x}=e^0=1$

4 0

2 years ago

| What is the product? (2x-1)(x +4) O 2x2-4 O 2x2+4 O 2x2 +78-4 O 2x2–7x-4

zubka84 [21]

The last one is that correct answer.
2x^2 +7x -4

3 0

2 years ago