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3 years ago

If sin yº = 7/q and tan yº = 7/r , what is the value of cos yº? (6 points)

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

4 0

Answer:

cos r/q is the answer,

Step-by-step explanation:

this can be seen when you elaborate sin=opposite/hypotenus and tan=opposite/adjacent by referring to the right angle triangle .

then, apply values of adjacent dan hypotenuse to cos

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For the equation, 2x + 6 = 3x - 10, what is the solution

geniusboy [140]

Answer: x=16

Step-by-step explanation:

$2x+6=3x-10$

Subtract 6 on both sides

$2x+6-6=3x-10-6$

$2x=3x-16$

Subtract 3x on both sides

$2x-3x=3x-16-3x$

$-x=-16$

Divide -1 on both sides

$\frac{-x}{-1}=\frac{-16}{-1}$

$x=16$

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3 years ago

Use the graph of the floor function to complete the statements. Over the interval , y = 2. For an input value of –1.5, the corre

nata0808 [166]

Answer:

over the interval [2,3)

for an input value of 1.5, the corresponding output value is -2

Step-by-step explanation:

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3 years ago

25a^2-30a+9 factor ?

CaHeK987 [17]

Factor 25a2−30a+9

25a2−30a+9

=(5a−3)(5a−3)

Answer:

(5a−3)(5a−3)

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4 years ago

Help with this, please!<br><br><br>It's about the volume of a cone

Lena [83]

Answer:

Step-by-step explanation:

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3 years ago

Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a

spin [16.1K]

Answer:

$Mean = 68.9$

$s^2 =18.1$ --- Variance

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

$x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5$

For 59 - 66

$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

For 75 - 82

$x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5$

For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

So, the table becomes:

$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

$Mean = \frac{2547.5}{37}$

$Mean = 68.9$ -- approximated

Solving (b): The sample variance:

This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

5 0

3 years ago