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viktelen [127]
2 years ago
14

Find all angles θbetween 0° and 2pi satisfying the given equation ? sin(θ)=2√2

Mathematics
1 answer:
Greeley [361]2 years ago
4 0
We are to determine <span>all angles between 0° and 2pi satisfying the given equation sin(θ) = 2√2. The answer is that no angles that would fit to the given equation. This is because the values of trigonometric functions such as cosine and sine do not go more than 1. Since </span><span>2√2 is equal to 2.83, then the answer is none.</span>
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Healthy Gym
ehidna [41]

Answer:

c=2d+5

Step-by-step explanation:

Let

d ------> the number of days

c -----> the cost

we know that

The cost is equal to the number of days multiplied by $2 per day plus the initial fee of $5

The linear equation in slope-intercept form is

c=md+b

where

m is the slope

b is the c-intercept (value of c when the value of d is equal to zero)

in this problem we have

m=$2 per day

b=$5

substitute

c=2d+5

8 0
3 years ago
Read 2 more answers
The midpoint of mn is point p at (-4,6). If point m is at (8,-2), what are the coordinates of point n?
VashaNatasha [74]
<u>x₁ + x₂</u>                             <u>y</u>₁ + <u>y</u>₂<u>
</u>     2      = M                        2     = M
<u>
x + 8</u>                                 <u>y + -2</u><u>
</u>    2    = -4                            2           =6
<u>multiply both sides by 2.

</u>x+8=-8                             y - 2 = 12
<u>x = -16</u>                             <u>y = 14
</u>
Final Answer: (-16, 14)



7 0
3 years ago
Read 2 more answers
What is the area of the figure?
wel

Answer: 448

Step-by-step explanation:

6 0
3 years ago
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Round to the nearest tenth.
laila [671]

Answer:

8.2 units

Step-by-step explanation:

Given that the only information for our triangle are 2 sides and 1 angle, we must use the Law of Cosines to find side BC

<u />

<u>Recall the Law of Cosines</u>

<u />a^2=b^2+c^2-2bc\cdot cos(A)

<u>Identify angles and sides</u>

<u />m\angle A=23^\circ\\a=BC=?\\b=AC=14\\c=AB=6

<u>Solve for side "a"</u>

<u />a^2=b^2+c^2-2bc\cdot cos(A)\\\\a^2=14^2+6^2-2(14)(6)\cdot cos(23^\circ)\\\\a^2=196+36-168cos(23^\circ)\\\\a^2=222-168cos(23^\circ)\\\\a=\sqrt{222-168cos(23^\circ)}\\ \\a\approx8.2

Therefore, the length of line segment BC is about 8.2 units

8 0
2 years ago
Last 2 questions please help
Jlenok [28]
4.
Minimum: 20
First Quartile: 44
Median: 51
Third Quartile: 57
Maximum: 86

Ascending order : 20 41 44 44 49 53 55 57 80 86
6 0
2 years ago
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