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Alborosie
3 years ago
11

If sin x = .42, what is cos x?

Mathematics
1 answer:
horrorfan [7]3 years ago
3 0
We\ know:\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\to\cos x=\sqrt{1-\sin^2x}\\\\\sin x=0.42\\subtitute\\\\\cos x=\sqrt{1-0.42^2}=\sqrt{1-0.1764}=\sqrt{0.8236}\approx0.91
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2 Evaluate the following expression when x = 2. x3 + 7x + 2
bogdanovich [222]

Answer:

B

Step-by-step explanation:

Since x is 2 wherever you see x put 2

solve x3 + x7+ 2

=(2)3 + (2)7 + 2

= you do the number outside the bracket multiplied by the number inside the bracket

like this

=(2 × 3) + (2 × 7) + 2

= 6+ 14 + 2

= 22

5 0
3 years ago
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liq [111]

Answer:

Angle 4 = 99 degrees

Step-by-step explanation:

Vertical Angles add up to 180

180-81= 99

4 0
3 years ago
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just olya [345]
53.2

40%=.4
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3 0
3 years ago
The table of values represents a function ​f(x).
TiliK225 [7]

Answer:

- The interval  [9, 10] is 1,375% greater than the interval [5, 8].

- The interval  [9, 10] is 14.75 times greater than the interval [5, 8].

Step-by-step explanation:

1. To solve this problem you must apply the following formula:

AverageRateOfChange=\frac{y_{2}-y_{1}}{x_{2}-y_{1}}

2. Let's calculate the average rate of change of each interval:

a) Interval [9,10]:

 =\frac{11,014-4,052}{10-9}=6,962

b) Interval [5,8]:

 =\frac{1,491-75}{8-5}=472

3. The difference is:

6,962-472=6,490

4. In percentage:

(\frac{6,962-472}{472})(100)=1,375%

5. You have that the interval [9,10] is 14.75 times greater than the interval [5,8], as you can see below:

\frac{6,962}{472}=14.75

3 0
4 years ago
Read 2 more answers
A bag contains 10 marbles: 3 are green, 2 are red, and 5 are blue. Ashley chooses a marble at random, and without putting it bac
Butoxors [25]

(1).probability of getting a red marble is.

total number of red marble is 2.

total number of marble is 10.

p(e) =  \frac{2}{10}  =  \frac{1}{5}

(2). probability of getting a blue marble is

total number of blue marble is 3.

total number of marble in bag is 9

p(e) =  \frac{3}{9}  =  \frac{1}{3}

6 0
3 years ago
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