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4 years ago

13

Help???????????????????

1 answer:

White raven [17]4 years ago

4 0

First get a common denominator. (3x*2)
x(x-8) -3(3x)
-------------------
3x^2

x^2 - 8x - 9x
------------------
3x^2

x^2 - 17x
--------------
3x^2
The first one is correct

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klasskru [66]

Answer:

This just explains what the temperature scale is.

Step-by-step explanation:

Have a look at the scale! :)

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3 years ago

(x + 23) + (2x + 4) = 180<br><br> Find the value of ¨x¨

Nata [24]

$(x+23)+(2x+4)=180$

$x+23+2x+4=180$

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$3x=180-27$

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4 years ago

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Which equation represents a line which is parallel to the line 3x-4y=283x−4y=28?

Fed [463]

Answer:

Step-by-step explanation:in standard form y= ax+b

-4y = -3x+28

4y =3x-28

or y= 3/4 x -7

So a= 3/4 = slope and -7 is y intercept ( parallel lines have same slope)

Equation 3 is

y= 3/4x-2

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Similarly a perpendicular line will have slope of - 1/a

For equation

y= ax+b

Perpendicular line is

y=( -1/a) x +c

5 0

3 years ago

What is the x -intercept of the graph y = 2 x - 14?<br><br> -14<br> -7<br> 0<br> 7

lidiya [134]

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3 years ago

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During a 90-day monsoon season, daily rainfall can be modeled by sinusoidal functions. A low of 4 inches of rainfall was recorde

Phoenix [80]

Answer:

$25.4\,days \leq t \leq 34.6\,days$

Step-by-step explanation:

Let consider the following model:

$c(t) = A \cdot \sin \left(\frac{2\pi \cdot t}{T} \right) + \bar c$

The average is given by the following formula:

$\bar c = \frac{c_{min}+c_{max}}{2}$

The maximum value is:

$c_{max} = 2 \cdot \bar c - c_{min}$

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$c_{max} = 12\,in$

Amplitude is:

$A = 12\,in - 8\,in$

$A = 4\,in$

The sine function has a periodicity of $2\pi$ , where is minimum is reached at $\theta = \frac{3\pi}{2}$ , when t = 30. The period of the cycle is:

$T = \frac{2\pi}{\frac{3}{2}\pi }\cdot (30)$

$T = 40$

The complete expression is:

$c(t) = 8\,in + 4\,in \cdot \sin \left(\frac{2\pi}{40}\cdot t \right)$

The times associated with $c = 5\,in$ are, respectively:

$8\,in + 4\,in \cdot \sin \left(\frac{2\pi}{40}\cdot t \right) = 5\,in$

$4\,in \cdot \sin \left(\frac{2\pi}{40}\cdot t \right) = -3\,in$

$\sin \left(\frac{2\pi}{40}\cdot t \right) = -0.75$

$t = \frac{40}{2\pi}\cdot \sin^{-1} (-0.75)$

Instants are, respectively:

$t_{1} \approx 25.4\,days$

$t_{2} \approx 34.6\,days$

Period is:

$25.4\,days \leq t \leq 34.6\,days$

8 0

3 years ago