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3 years ago

Trigonometry can anyone help me out

Mathematics

2 answers:

Reil [10]3 years ago

4 0

Answer:

y = 11

x = 11 root 3

$x = 11\sqrt{3}$

$y = 11$

Step-by-step explanation:

Cos 60 =

$\frac{y}{22} \\ 0.5 = \frac{y}{22} \\ 0.5 \times 22 = y \\ 11 = y$

$for x \\ \sin(60) = \frac{x}{22} \\ \frac{ \sqrt{3} }{2} = \frac{x}{22} \\ \frac{ \sqrt{3} }{2} \times 22 = x \\ 11 \sqrt{3 } = x$

Bond [772]3 years ago

3 0

Answer:

y = 11

x = 11 root \3

Step-by-step explanation:

You might be interested in

2 (01.05)

lord [1]

Answer:

y +2 = 4(x -3)

Step-by-step explanation:

Point-slope form of the equation of a line with slope m through point (h, k) is ...

y -k = m(x -h)

For m = 4 and (h, k) = (3, -2), the equation is ...

y +2 = 4(x -3) . . . . . matches the first choice

3 0

3 years ago

Please help me with this thanks please let me know if my answer is incorrect

sergey [27]

I just graphed the table next to the graph of g and compared the two.

The y-intercept of f is (0,8) and the y-intercept of g is (0, -2). So f is GREATER THAN the y-intercept of g.

Now to do the average rate of change, I see from the interval [-6,-3] that f is decreasing and for g it is increasing. Thus f is LESS THAN the average rate of change of g.

5 0

2 years ago

Xis directly proportional to y.

OleMash [197]

X = k x y
21 = k x 3
k = 7
x = 7y

x = 7(10)
x = 70

6 0

3 years ago

The volume of the pyramid shown in the figure is cubic centimeters. If the slant height of the pyramid increases by 4 centimeter

Nostrana [21]

Answer:

Part a) The volume of the original pyramid is $15\ cm^{3}$

Part b) The volume of the pyramid increases by $6\ cm^{3}$

Step-by-step explanation:

The complete question is

Part 1) The volume of the pyramid shown in the figure is 9,15,21, or 63 cubic centimeters?. Part 2) If the slant height of the pyramid increases by 4 centimeters and its height increases by 2 centimeters, the volume of the pyramid increases by 6,9,12 or 21 cubic centimeters?

we know that

The volume of the pyramid is equal to

$V=\frac{1}{3}Bh$