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8090 [49]
3 years ago
12

Only 15 and 16 are the one I'm confused on

Mathematics
1 answer:
PtichkaEL [24]3 years ago
8 0
Okay so you see where it says f(2) you plug that into the equation so then you would have f(2)=x2+3x-2 and then number 16 is the same. 
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A right cylinder and an oblique cylinder have the same radius and the same height. How do the volumes of the two
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Answer:

  A.  The volumes are the same, based on Cavalieri's principle.

Step-by-step explanation:

Cavalieri's principle tells us the volumes of solids will be identical if their cross sectional areas are identical at every height.

<h3>Application</h3>

A right cylinder and an oblique cylinder of the same height and radius will both have circular cross sections of the given radius at any height. Since the radius is the same, the area of the circle is the same. Hence the requirements of Cavalieri's principle are met, and the cylinders have the same volume.

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2 years ago
If cos(θ)= -2/3 and θ lies in quadrant II, find the value of cotθ
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\text{Given that,}\\\\\cos \theta = - \dfrac 23\\\\\implies \cos^2 \theta  = \dfrac 49\\\\\implies 1- \sin^2 \theta = \dfrac 49\\\\\implies \sin^2 \theta =1-\dfrac 49\\\\\implies \sin^2 \theta =\dfrac 59\\\\\\\text{Now,}\\\\\cot^2 \theta = \dfrac{\cos^2 \theta}{\sin^2 \theta} \\\\\\\implies \cot^2 \theta = \dfrac{\tfrac 49}{\tfrac 59}\\\\\\\implies \cot^2 \theta = \dfrac 49 \times \dfrac 95\\\\\\\implies \cot^2 \theta = \dfrac 45\\\\\\

\implies \cot \theta = \pm\sqrt{\dfrac 45} =  \pm \dfrac 2{\sqrt 5}\\\\\\\text{Since}~ \theta ~ \text{lies in the second quadrant,}  \cot \theta ~\text{will be negative.}\\\\\text{Hence,}~ \cot \theta = - \dfrac{2}{\sqrt 5}

6 0
2 years ago
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Step-by-step explanation:

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Figure ABCD has vertices A(−4, 1), B(2, 1), C(2, −5), and D(−4, −3). What is the area of Figure ABCD?
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Check the picture below.

so, notice, the composite is just a 4x6 rectangle, and a triangle whose base is 6 and altitude is 2.  So, we can simply get the area of each figure and sum them up.

\bf \stackrel{\textit{area of rectangle}}{(4\cdot 6)}~~~~+~~~~\stackrel{\textit{area of triangle}}{\cfrac{1}{2}(6)(2)}

5 0
3 years ago
Can someone please help me I reallly need help <br> please help
Serggg [28]

Answer:

c = 38b

Step-by-step explanation:

b = number of boxes

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c = 38b is your answer.

~

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