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4 years ago

PLS HELP ME WITH THIS! (The “GIVEN” one) THANK YOU (Random answers gets moderated.)

Mathematics

1 answer:

kvasek [131]4 years ago

7 0

I can't say i did the problem but from a general format. 0 </= x </= 5 are what is called and interval [0 to 5] those are your x values, your function is of course the y=-x+3 plug in the x values, which will give you the y. graph it, using the x and corresponding y. do the same for the 2nd portion. and you already know your domain (x) as b already tells you. 0,1,2,3,4,5..

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Art [367]

Given:
Parallelogram length = 11 cm
triangle atop the parallelogram : short leg = 7 cm
Right triangle inside the parallelogram : long leg = 9cm

Area of parallelogram = base * height
A = 11cm * 9cm
A = 99 cm²

Area of a triangle atop the parallelogram:
A = ab/2
A = (11cm * 7cm)/2
A = 77 cm² / 2
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Total area of the figure
99 cm² + 38.5 cm² = 137.5 cm²

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1. For y = 4x^2 + 9 − 5x^4 − x^3 a. Rewrite the equation in standard form: b. Identify the degree: c. Describe the end behavior:

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Help me solve them plsss...both of the questions :)

zaharov [31]

$\bold{\text{Answer:}\quad \dfrac{\sqrt2}{144}}$

<u>Step-by-step explanation:</u>

When you plug in 3 directly to the equation, you get 0/0

Since it is indeterminate, you have to use L'Hopital's Rule.

That means that you find the limit of the DERIVATIVE of the numerator and the DERIVATIVE of the denominator.

$\dfrac{d}{dx}(\sqrt{5+\sqrt{2x+3}}-2\sqrt2)\quad = \dfrac{1}{2\sqrt{5+\sqrt{2x+3}}(\sqrt{2x+3})}\\\\\\f'(3)\ \text{for the numerator}\ =\dfrac{1}{12\sqrt2}\\\\\\\\\dfrac{d}{dx}(x^2-9) = 2x\\\\\\f'(3) \text{for the denominator}\quad = 6\\\\\\\dfrac{f'(3)\ \text{numerator}}{f'(3)\ \text{denominator}}\quad = \dfrac{\dfrac{1}{12\sqrt2}}{6}\quad = \dfrac{1}{72\sqrt2}\quad \rightarrow \large\boxed{\dfrac{\sqrt2}{144}}$

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3 years ago

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