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Andrews [41]
3 years ago
15

Y = sec θ tan θ differentiate

Mathematics
1 answer:
Leviafan [203]3 years ago
6 0
Y = sec t tan t = 1/ cos t * sin t / cos t = sin t / cos² t
y`= \frac{cos t cos ^{2} t-2 cost*(-sin t)*sint}{cos^{4} t}= \\  =\frac{cos ^{3} t+2costsin^{2}t }{cos ^{4}t }= \\ = \frac{cos ^{2} t+2sin ^{2} t}{cos ^{3} t}= \\ = \frac{1-sin ^{2}t+2sin ^{2} t }{cos ^{3}t }= \\ = \frac{1+sin ^{2} t}{cos ^{3}t }
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4-(-3) as an addition problem and find the difference.
Naddika [18.5K]

\huge\text{Hey there!}

\huge\boxed{\text{Guide \#1}\downarrow}

\bullet \huge\textsf{ Negative (-) \& negative (-) = positive (+)}

\bullet \huge\textsf{ Positive (+) \& negative (-) = negative (-)}

\bullet\huge\textsf{ Positive (+)  \& positive (+) = positive (+)}

\bullet\huge\textsf{ Negative (-) \& positive (+)  = negative (-)}

\huge\boxed{\frak{Guide\  \#2}\downarrow}

\bullet \huge\textsf{ Positives are on the RIGHT side of the}\\\huge\textsf{ number line}

\bullet\huge\textsf{ Negatives are on the LEFT side of the}\\\huge\textsf{ number line}

\huge\boxed{\bf Guide\  \#3\downarrow}}

\bullet\huge\textsf{ Positives are ABOVE 0}

\bullet\huge\textsf{ Negatives are BELOW 0}

\huge\boxed{Answering \ your \ question: }

\huge\boxed{\mathsf{4 - (-3)}}

\huge\boxed{\mathsf{= 4 + 3}}

\star \large\textsf{ You start at 4 and go up 3 spaces to the right because we are dealing with } \\\textsf{positive numbers on both right and left side }\star\huge\boxed{\mathsf{= 4 + 3}}

\huge\boxed{\mathsf{= 7}}

\huge\boxed{\frak{\bold{Therefore, \ your \ \underline{\underline{answer \  is: \boxed{\frak 7}}}}}}\huge\checkmark

\huge\text{Good luck on your assignment \& enjoy your day!}

<h2>~\huge\boxed{\rm{Amphitrite1040:)}}</h2>
6 0
3 years ago
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The minimum of the graph of a quadratic function is located at (–1, 2). The point (2, 20) is also on the parabola. Which functio
ss7ja [257]
The min or max of a parabola/quadratic function is the vertex

for
y=a(x-h)²+k
the vertex is (h,k)

so
vertex/min is at (-1,2)
h=-1
k=2

y=a(x-(-1))²+2
y=a(x+1)²+2
find a
given, (2,20) is on the graph

20=a(2+1)²+2
20=a(3)²+2
20=9a+2
minus 2 both sides
18=9a
divide by 9
2=a


y=2(x+1)²+2 is da equation

3rd one
f(x)=2(x+1)²+2
5 0
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