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Natalka [10]
3 years ago
6

Together, teammates Pedro and Ricky got 2689 base hits last season. Pedro had 285 more hits than Ricky. How many hits did each p

layer​ have?
Mathematics
1 answer:
STALIN [3.7K]3 years ago
7 0
Pedro had 1487 hits
Ricky had 1202 hits

2688/2=1344.5
285/2=142.5
1344.5-142.5=1202 (Ricky)
1344.5+142.5=1487 (Pedro)
1487-1202=285
1487+1202=2689
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A guitar string is plucked at a distance of 0.6 centimeters above its resting position and then released, causing vibration. The
liubo4ka [24]

Answer:

The equation that represents the motion of the string is given by:

y =Ae^{-kt}\cos(2\pi ft)     .....[1] where t represents the time in second.

Given that: A = 0.6 cm (distance above its resting position) , k = 1.8(damping constant) and frequency(f) = 105 cycles per second.

Substitute the given values in [1] we get;

y =0.6e^{-1.8t}\cos(2\pi 105t)  

or

y =0.6e^{-1.8t}\cos(210\pi t)  

(a)

The trigonometric function that models the motion of the string is given by:

y =0.6e^{-1.8t}\cos(210\pi t)

(b)

Determine the amount of time t that it takes the string to be damped so that -0.24\leq y \leq0.24

Using graphing calculator for the equation

y =0.6e^{-1.8x}\cos(210\pi x)

let x = t (time in sec)  

Graph as shown below in the attachment:

we get:

the amount of time t that it takes the string to be damped so that -0.24\leq y \leq0.24 is, 0.5 sec


4 0
3 years ago
B
Tomtit [17]
This would be 525 / 0.70  = $750 Answer
8 0
3 years ago
1. cot x sec4x = cot x + 2 tan x + tan3x
Mars2501 [29]
1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
    cot(x)sec⁴(x)            cot(x)sec⁴(x)
                   0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
                   0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
                   0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
                   0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
                   0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
                   0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
                   0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
                   0 = cos⁴(x)(1 + tan²(x))²
                   0 = cos⁴(x)        or         0 = (1 + tan²(x))²
                ⁴√0 = ⁴√cos⁴(x)      or      √0 = (√1 + tan²(x))²
                   0 = cos(x)         or         0 = 1 + tan²(x)
         cos⁻¹(0) = cos⁻¹(cos(x))    or   -1 = tan²(x)
                 90 = x           or            √-1 = √tan²(x)
                                                         i = tan(x)
                                                      (No Solution)

2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
              sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
   sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
                               sin²(x) - cos²(x) = sin²(x) - cos²(x)
                                         + cos²(x)              + cos²(x)
                                             sin²(x) = sin²(x)
                                           - sin²(x)  - sin²(x)
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3. 1 + sec²(x)sin²(x) = sec²(x)
           sec²(x)             sec²(x)
      cos²(x) + sin²(x) = 1
                    cos²(x) = 1 - sin²(x)
                  √cos²(x) = √(1 - sin²(x))
                     cos(x) = √(1 - sin²(x))
               cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
                                 x = 0

4. -tan²(x) + sec²(x) = 1
               -1               -1
      tan²(x) - sec²(x) = -1
                    tan²(x) = -1 + sec²
                  √tan²(x) = √(-1 + sec²(x))
                     tan(x) = √(-1 + sec²(x))
            tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
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emmainna [20.7K]

Answer:

Ummmm no thanks. I am not going there.

Step-by-step explanation:

6 0
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dmitriy555 [2]

Answer:

98

Step-by-step explanation:

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