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GenaCL600 [577]

2 years ago

9

X = y - 3

1 answer:

Leviafan [203]2 years ago

3 0

The answer is A. (1, 4), because when the values are substituted in to the equations, you get 1 = 4 - 3 and 1 + 12 = 13, which are both correct. I hope this helps!

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Write the equation of a line, in slope intercept form, that has a slope of 1 and passes through the point (2, 4).

Fudgin [204]

The answer is y=x+2 because you can put it into point slope form

y-4=1(x-2)simplify

y-4=x-2

y=x+2

4 0

2 years ago

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Which of the following describes the transformation of g(x) = -(2)x+4 -2 from the parent function f(x)= 2^x

Nataly [62]

Ok, so what happend was 4 was added to every x, the whole thing was multiplied by -1 and -2 was added to the whole function

so
adding 4 to every x shifts the graph to the left 4 units
multiplying the whole function by -1 reflects it over the x axis
adding -2 to the whole function moves it down2 units

so that would be the first option

4 0

2 years ago

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Find dy/dx given that y = sin x / 1 + cos x

kobusy [5.1K]

Answer:

$\frac{1}{1 + \cos(x) }$

Step-by-step explanation:

$y = \frac{ \sin(x) }{1 + \cos(x) }$

<u>differentiating numerator wrt x :-</u>

(sinx)' = cos x

<u>differentiating denominator wrt x :- </u>

(1 + cos x)' = (cosx)' = - sinx

Let's say the denominator was "v" and the numerator was "u"

$(\frac{u}{v} )' = \frac{v. \: (u)' - u.(v)' }{ {v}^{2} }$

here,

since u is the numerator u= sinx and u = cos x
v(denominator) = 1 + cos x; v' = - sinx

$= \frac{((1 + \cos \: x) \cos \: x )- (\sin \: x. ( - \sin \: x) ) }{( {1 + \cos(x)) }^{2} }$

$= \frac{ \cos(x) + \cos {}^{2} (x) + \sin {}^{2} (x) }{(1 + \cos \: x) {}^{2} }$

since cos²x + sin²x = 1

$= \frac{ \cos \: x + 1}{(1 + \cos \: x) {}^{2} }$

diving numerator and denominator by 1 + cos x

$= \frac{1}{1 + \cos(x) }$

6 0

2 years ago

Solve x - 6 = -12 A) -18 B) -6 C) 2 D) 6

asambeis [7]

The answer would be -6 or letter B because
$x - 6 = - 12 \\ x - 6 + 6 = - 12 + 6 \\ x = - 6$

8 0

2 years ago

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An airplane needs to reach a velocity of 203.0 km/h to take off. On a 2000 m runway, what is the minimum acceleration necessary

Aliun [14]

Answer:

1.59 m/s²

Step-by-step explanation:

Acceleration = change in Velocity over time. Mathematically it is represented as Δv/Δt

Given values in the question:

Final velocity 203.0 km/h which is equal to (203/3.6)m/s = 56.39 m/s
distance 2000 m
Initial velocity is 0 m/s

<em>To use the equation for acceleration time is needed.</em>

<em>velocity= distance/time</em>

(203/3.6) = 2000/time <em>mltiplying both sides by time</em>

(203/3.6) x time = 2000 <em>making time subject of the formula</em>

time = 2000/(203/3.6)

time = 35.47 seconds

Substituting into the acceleration equation

acceleration =Δv/Δt

=(56.39- 0)/35.47

= 56.39/35.47

=1.59 m/s²

6 0

2 years ago