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[flierfly]

I HAVE A QUESTION!
TYPE OF QUESTION:Math
YOUR QUESTION: y= 2x² x=-5

[LostInTheEcho]

I HAVE A QUESTION!
TYPE OF QUESTION:Math
YOUR QUESTION: y= 2x² x=-5

I'm not 100% what you're supposed to calculate, but I'm guessing it's the value of y when x=-5.

You start by substituting x with -5 in the equation
y=2*(-5)^2
You take the power of 2 before multiplying with 2 (since that is the order of operations), and two negative numbers multiplied with each other is positive [ (-5)^2 = (-5)*(-5) ]
y=2*25
And two times 25 is 50 so
y=50 when x=-5

Hope that answers your question

[Blumenkranz]

I HAVE A QUESTION!
TYPE OF QUESTION: Trigonometry
YOUR QUESTION: Use the given information and a calculator to find θ to the nearest tenth of a degree if 0 < θ < 360°.
tan θ = 0.5890 with θ in Quadrant III.

[LostInTheEcho]

I HAVE A QUESTION!
TYPE OF QUESTION: Trigonometry
YOUR QUESTION: Use the given information and a calculator to find θ to the nearest tenth of a degree if 0 < θ < 360°.
tan θ = 0.5890 with θ in Quadrant III.

On the calculator you're supposed to use the arctan (tan^-1) function since that is the opposite of using tan.

The third quadrant is between 180 and 270 degrees as can be seen in the image below


The tangent is a peridic function that that repeat with the intervall 180 degrees. So to find the angle θ in the third quadrant you need to add 180° to the answer you get using arctan.

How to find your answer:
θ = arctan (0.5890) + 180°

And then round the answer
(Make sure your calculator is set to degrees and not radians)

Hope this helps!

[Blumenkranz]

@LostInTheEcho, that makes sense, thank you for explaining it!

[dovah]

I HAVE A QUESTION!
TYPE OF QUESTION: calculus
YOUR QUESTION:
which of the following is an equation for the tangent line to f(x) = 9-x^2 at x=2?

a) y=1/4x+9/2
b) y=-4x+13
c) y=-4x=3
d) y=4x-3
e) y=4x+13

[Piefan]

I HAVE A QUESTION!
TYPE OF QUESTION: calculus
YOUR QUESTION:
which of the following is an equation for the tangent line to f(x) = 9-x^2 at x=2?

a) y=1/4x+9/2
b) y=-4x+13
c) y=-4x=3
d) y=4x-3
e) y=4x+13

Differentiate f(x), and you get -2x.

Hence, when x=2, the gradient of the graph is (-2)(2) = -4

y=mx+c (or whatever equation you were taught). So the only possible answers are (b) and (c).

Looking at the original equation again, y=9-x^2. So when x=2, y=5.

So now we need to figure out which line passes through that point, right? Let's check it against both (b) and (c).

(b): 5=-4(2)+13
5=-8+13

(c): 5=-4(2)=3

C is blatantly untrue. B is correct.

[dovah]

snip

Differentiate f(x), and you get -2x.

Hence, when x=2, the gradient of the graph is (-2)(2) = -4

y=mx+c (or whatever equation you were taught). So the only possible answers are (b) and (c).

Looking at the original equation again, y=9-x^2. So when x=2, y=5.

So now we need to figure out which line passes through that point, right? Let's check it against both (b) and (c).

(b): 5=-4(2)+13
5=-8+13

(c): 5=-4(2)=3

C is blatantly untrue. B is correct.

oops i actually just made an error while typing it out, c) is supposed to say y=-4x-3. and thank u, i think i'm getting it

[airam]

I HAVE A QUESTION!
TYPE OF QUESTION: Kinematics
YOUR QUESTION:

The lift in a tall building passes the 50th floor with a velocity of -8 m/s and an acceleration of ⅑(t-5) m/s/s (m/s^2). If each floor spans a height of 6 metres, find at which floor the lift will stop.

note: t = time in seconds

Any help greatly appreciated. I do have the answer but I don't understand the literacy of the question/the steps you're supposed to take.

[Jaskiest]

I HAVE A QUESTION!
TYPE OF QUESTION: Kinematics
YOUR QUESTION:

The lift in a tall building passes the 50th floor with a velocity of -8 m/s and an acceleration of ⅑(t-5) m/s/s (m/s^2). If each floor spans a height of 6 metres, find at which floor the lift will stop.

note: t = time in seconds

Any help greatly appreciated. I do have the answer but I don't understand the literacy of the question/the steps you're supposed to take.

Let me know if you need clarification anywhere! :)