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4 years ago

Question 3

Mathematics

1 answer:

Papessa [141]4 years ago

4 0

Answer:

(a) The next 3 terms in the sequence are 17, -31, and 65

(b) The 3 terms that came before -7 are 2, -1, and 5

Step-by-step explanation:

a. To determine the next 3 terms in the sequence after -7,

From the question, the sequence follows the rule: multiply the previous number by-2 and add 3.

Then, for the next number after -7, we will get

-7 × -2 = 14; 14 + 3 = 17

For the second number after -7, we will get

17 × -2 = -34; -34 + 3 = -31

For the third number after -7, we will get

-31 × -2 = 62; 62 + 3 = 65

Hence, the next 3 terms are 17, -31, and 65

(b) For the 3 terms that came before -7 in the sequence, the rule will be reversed, that is: subtract 3 from the immediate number and divide by -2

Hence, the previous number to -7 (that is, the third term in the sequence)

-7 -3 = -10; -10 ÷ -2 = 5

Then, the second term will be

5 - 3 = 2; 2 ÷ -2 = -1

and the first term will be

-1 -3 = -4; -4 ÷ -2 = 2

Hence, the 3 terms that came before -7 are 2, -1, and 5

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Blank divided by 7>800

djverab [1.8K]

Hi there

Replace the blank space by x

so x/7 > 800
All you have to do is cross-multiply

x > 7*800
x> 5600

I hope that's help !

8 0

3 years ago

Read 2 more answers

Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction p

kirza4 [7]

The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

<h3>How to convert sine of an angle to some angle of cosine?</h3>

We can use the fact that:

$\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)$

to convert the sine to cosine.

<h3>Which trigonometric functions are positive in which quadrant?</h3>

In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
In second quadrant(π/2 < θ < π), only sin and cosec are positive.
In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

$y= 3\cos(2(x + \pi/2)) - 2$

The options are:

$y= 3\sin(2(x + \pi/4)) - 2$
$y= -3\sin(2(x + \pi/4)) - 2$
$y= 3\cos(2(x + \pi/4)) - 2$
$y= -3\cos(2(x + \pi/2)) - 2$

Checking all the options one by one:

Option 1: $y= 3\sin(2(x + \pi/4)) - 2$

$y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2$

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

Option 2: $y= -3\sin(2(x + \pi/4)) - 2$

This option if would be true, then from option 1 and this option, we'd get:
$-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0$