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Wewaii [24]
2 years ago
14

the table above shows the values of functions f and h and their derivatives at x=4 and x=8. what is the value of d/dx (f(h(x)))

at x=4?​

Mathematics
1 answer:
solmaris [256]2 years ago
8 0
The answer is C. You have to use chain rule

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Find the vertices and foci of the hyperbola. 9x2 − y2 − 36x − 4y + 23 = 0
Xelga [282]
Hey there, hope I can help!

NOTE: Look at the image/images for useful tips
\left(h+c,\:k\right),\:\left(h-c,\:k\right)

\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1\:\mathrm{\:is\:the\:standard\:equation\:for\:a\:right-left\:facing:H}
with the center of (h, k), semi-axis a and semi-conjugate - axis b.
NOTE: H = hyperbola

9x^2-y^2-36x-4y+23=0 \ \textgreater \  \mathrm{Subtract\:}23\mathrm{\:from\:both\:sides}
9x^2-36x-4y-y^2=-23

\mathrm{Factor\:out\:coefficient\:of\:square\:terms}
9\left(x^2-4x\right)-\left(y^2+4y\right)=-23

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1
\frac{1}{1}\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
\frac{1}{1}\left(x^2-4x+4\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)

\mathrm{Convert\:to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y+4\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)

\mathrm{Convert\:to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)

\mathrm{Refine\:}-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right) \ \textgreater \  \frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=1 \ \textgreater \  Refine
\frac{\left(x-2\right)^2}{1}-\frac{\left(y+2\right)^2}{9}=1

Now rewrite in hyperbola standardform
\frac{\left(x-2\right)^2}{1^2}-\frac{\left(y-\left(-2\right)\right)^2}{3^2}=1

\mathrm{Therefore\:Hyperbola\:properties\:are:}\left(h,\:k\right)=\left(2,\:-2\right),\:a=1,\:b=3
\left(2+c,\:-2\right),\:\left(2-c,\:-2\right)

Now we must compute c
\sqrt{1^2+3^2} \ \textgreater \  \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \  1^2 = 1 \ \textgreater \  \sqrt{1+3^2}

3^2 = 9 \ \textgreater \  \sqrt{1+9} \ \textgreater \  \sqrt{10}

Therefore the hyperbola foci is at \left(2+\sqrt{10},\:-2\right),\:\left(2-\sqrt{10},\:-2\right)

For the vertices we have \left(2+1,\:-2\right),\:\left(2-1,\:-2\right)

Simply refine it
\left(3,\:-2\right),\:\left(1,\:-2\right)
Therefore the listed coordinates above are our vertices

Hope this helps!

8 0
3 years ago
Using the geometric mean and Pythagorean theorem, calculate the values of the missing sides. Round your answers to the thousandt
Pachacha [2.7K]

Answer:

a = 9.849

b = 20.25

c = 491.03

Step-by-step explanation:

By using Pythagoras theorem in the right triangle BDC,

(Hypotenuse)² = (Leg 1)² + (Leg 2)²

BC² = BD² + DC²

a² = 9² + 4²

a = \sqrt{(81+16)}

a = \sqrt{97}

a = 9.8489

a ≈ 9.849 units

By mean proportional theorem,

\frac{\text{DC}}{\text{BD}}=\frac{\text{BD}}{\text{AD}}

AD × DC = BD²

b × 4 = 9²

b = \frac{81}{4}

b = 20.25 units

BY Pythagoras theorem in ΔADB,

AB² = AD² + BD²

c² = b² + 9²

c² = (20.25)² + 9²

c² = 410.0625 + 81

c = 491.0625

c = 491. 063 units

6 0
3 years ago
Find two integers whose sum is -10 and product is 16<br> TWO INTERGES PLEASE
Lady bird [3.3K]

Answer:

-2 and -8

Step-by-step explanation:

-2 + (-8) = -10

-2 * -8 = 16

8 0
3 years ago
Find the measure of the missing angle.
melamori03 [73]

Answer:

It's 155

Step-by-step explanation:

Since the whole angle is 180

And u got 25

So 180-25=155

4 0
3 years ago
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PIT_PIT [208]
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It makes sense. At most means equal or less than. The 10 per hour has a variable.
8 0
2 years ago
Read 2 more answers
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