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miv72 [106K]
3 years ago
9

Help with 4 and 5 please!!

Mathematics
1 answer:
Ghella [55]3 years ago
4 0
Question 4
The magnitude;
Using Pythagoras theorem,
 (-200)² + (-530)² = 320900
                 Length = √320900
                              =  566.5 mi
To get the angle
Tan θ = opposite/adjacent
          = 200/530
          =  0.3774
       θ = tan^-1 (0.3774)
          =  20.67
          ≈ 21°
The direction from the Cartesian plane is south of west.
Therefore, the magnitude and the direction will be;
 About 566.5 mi, 21° south of west

Question 5.
To get the resultant of two vectors we just add the two vectors given.
This involves adding the corresponding values.
Thus, for <-6,5> and <6,-5>
Resultant vector = <(-6+6),(5+-5>
                           = <0,0>
 
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#6 from the paper please
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Answer:

EK = 9

Step-by-step explanation:

x + 11 = -5 + 2x -5 + 2x

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11 = 3x - 10

3x = 21

x = 7

EK =-5 + 2 × 7

EK = 14 - 5

EK = 9

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If 9 pairs of jeans cost $121.50 find the cost of 5 pairs
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Which function will have a y-intercept at –1 and an amplitude of 2?
Angelina_Jolie [31]

Answer:

Options B and D.

Step-by-step explanation:

The general form of sine function

h(x)=A\sin (Bx+C)+D

where, |A| is amplitude, \frac{2\pi}{B} is period, \frac{-C}{B} is phase shift and D is y-intercept.

The general form of cosine function

h(x)=A\cos (Bx+C)+D

where, |A| is amplitude, \frac{2\pi}{B} is period, \frac{-C}{B} is phase shift and D is y-intercept.

In function, f(x)=-\sin x-1

Amplitude : |A|=1

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In function, f(x)=-\cos x

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In function, f(x)=-2\cos x-1

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4 0
4 years ago
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Find the measure of RST.
almond37 [142]

Answer:

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Step-by-step explanation:

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BRIANLY FOR THE BEST EXPLANATION AND ANSWER!
Kryger [21]

Answer:

C ≈ 14.18

Step-by-step explanation:

<u><em>Question Given:</em></u>

A circle has an area of 16mm². Find it's circumference.

<u><em>Formula:</em></u>

C = 2√πA

<u><em>Solve:</em></u>

C = 2√πA

C = 2√(3.14 × 16)

C = 2√50.24

C = 14.1760361173

C ≈ 14.18

<em>Hence, Circumference is approximately 14.18.</em>

<em />

<u>~Kavinsky</u>

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