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3 years ago

Please help me out giving brainliest

Mathematics

2 answers:

lara31 [8.8K]3 years ago

7 0

I’m not sure how to solve the first one but the second one (c) is 24. You have to see the pattern between 15 and 45 and then do that to the denominators. The pattern is multiplying by 3 (15x3 =45) so multiply 8 by 3 and you get 24. Hope this helps and if not, please message me for further help.

7nadin3 [17]3 years ago

6 0

For the first one, cross multiply 15 and c which is 15c and cross multiply 8 and 45 which is 360, so u get 15c = 360 and divide 15 c by 15 and 360 by 15 also then the answer would be c = 24

For the second one it’s basically the same thing, cross multiply x and 12 which is 12x and 8 and 3 which is 24, then you get 12x = 24. Divide both sides by 12 and the answer would be x = 2

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A landscaper is comparing the costs of potted plants from two nurseries. Nursery P charges $15 per plant plus a $50 delivery fee

photoshop1234 [79]

Answer:

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3 years ago

What is the inverse of the function 9y - 6 = 3x ?

IgorC [24]

Answer:

Option (2)

Step-by-step explanation:

Given function is,

9y - 6 = 3x

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To find the inverse of the given function,

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x = $\frac{3y+6}{9}$

2). Now we have to solve this equation for the value of y.

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Therefore, inverse of the given function is y = (3x - 2)

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3 years ago

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:
$-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0$