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Ronch [10]
3 years ago
11

50 POINTS! What is the value of arcsin(−1/2) in degrees?

Mathematics
2 answers:
Veronika [31]3 years ago
4 0

Answer:

-\frac{\pi }{6}

Step-by-step explanation:

arcsin(−1/2) = -arcsin(1/2)

In the table of common values,

-arcsin of 1/2 = π/6 = -π/6

Note: Memorize the results of the common values.

Stella [2.4K]3 years ago
3 0

Answer: 330°

Step-by-step explanation:

arcsin means sin^-1

So, arcsin(−1/2) is sin^-1(−1/2).

Now,

Let y = sin^-1(−1/2)

y = -30°

If range is between 0° and 360° then,

y = 360° - 30°

y = 330°

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Number 5 but if u can do number 6 that would be great ✋
mamaluj [8]

Answer:

5: 45 miles

6: y=20x

Step-by-step explanation:

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3 years ago
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ANSWER ASAP |What is the measure of angle x?
Marina86 [1]

Answer:

x = 83°

Step-by-step explanation:

41, 56 and the missing angle = 180° ( form a straight angle )

missing angle = 180° - (41 + 56)° = 180° - 97° = 83°

The angle x and the missing angle are vertically opposite and congruent

Hence x = 83°

8 0
3 years ago
1-5x + 3y 45<br>2x+3y = 24​
Norma-Jean [14]

Answer:

The answer is 1-5x+135y

Step-by-step explanation: multiply the numbers 3 and 45 which give you 135

3 0
3 years ago
Which statements are true?
Murrr4er [49]

Answer:

All squares are parallelograms, and all rectangles are quadrilaterals are both true

Step-by-step explanation:

4 0
3 years ago
If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f
Anni [7]

Answer:

Part A) Annual \$66,480.95  

Part B) Semiannual \$66,862.38  

Part C) Monthly \$67,195.44  

Part D) Daily \$67,261.54  

Step-by-step explanation:

we know that

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

so

Part A) Annual

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=1  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{1})^{1*5}  

A=\$47,400(1.07)^{5}  

A=\$66,480.95  

Part B) Semiannual

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=2  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{2})^{2*5}  

A=\$47,400(1.035)^{10}  

A=\$66,862.38  

Part C) Monthly

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=12  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{12})^{12*5}  

A=\$47,400(1.0058)^{60}  

A=\$67,195.44  

Part D) Daily

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=365  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{365})^{365*5}  

A=\$47,400(1.0002)^{1,825}  

A=\$67,261.54  

8 0
3 years ago
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