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GaryK [48]
3 years ago
15

F(x) = x^-1, x=.9. Use linear approximation to estimate answer

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
8 0
L(x) = f'(a)(x - a)
f(x) = x^(-1), f'(x) = -x^(-2)
Let a = 1, f'(a) = -(1)^(-2) = -1
L(0.9) = f'(1)(0.9 - 1) = -1(-0.1) = 0.1

Therefore, linear approximation of f(x) = x^(-1) for x = 0.9 is 0.1
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Use the given information to find the exact value of the trigonometric function
eimsori [14]
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Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

The answer is none of the choices

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5 0
3 years ago
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ludmilkaskok [199]

32.3 is a decimal number in which 3 is the digit at the tenth place.

Mixed number contains a whole number and a fraction part.

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\frac{3}{10} is the fraction in the simplest form.

For this reason, 32.3 expressed as a mixed number in the simplest form is 32\frac{3}{10}.

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dezoksy [38]

Answer:

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6 0
3 years ago
Triangle ABC is an isosceles right triangle inscribed in a circle. The center of the circle is point D and the diameter of the c
Paha777 [63]

Answer:

* AD is congruent to DC and BD <em>true</em>

* m∠B = 90° <em>true</em>

<u>* The measure of arc AC is equal to the measure of arc AB </u><u><em>not be true</em></u><em> ( The right answer  )</em>

* The measure of arc AB is equal to measure of arc BC <em>true</em>

Step-by-step explanation:

∵ D is the center of the circle and A , B and C are points on the circle

∴ AD , DB and DC are radii on the circle D

∴ AD ≡ DC ≡ DB

∵ AC passing through point D which is the center of the circle

∴ AC is the diameter of the circle D

∵ ∠B is opposite to the diameter of the circle and vertex B lies on the circle

∵ ∠B is an inscribed angle and  ∠ADC  is a central angle subtended by the same arc AC

∴m∠B = half m∠ADC

∵ m∠ADC = 180°

∴ m∠B = 90°

∵ The measure of arc AC = 180°

∵ ΔABC is isosceles and m∠B = 90°

∴ m∠BAC = m∠BCA = (180° - 90°) ÷ 2 = 45°

∵ ∠ACB is an inscribed angle subtended by arc AB

∴ m∠ACB = half measure of arc AB

∵ The measure of arc AB = 45° × 2 = 90°

∴ The measure of arc AC ≠ the measure of arc AB

∵ Δ ABC is an isosceles triangle and m∠B = 90°

∴ AB = BC

∵ AB subtended by arc AB

∵ BC subtended by arc BC

∴ The length of arc AB = the length of arc BC

∵ If two arcs are equal in length, then they will be equal in measure

∴ The measure of arc AB is equal to the measure of arc BC

3 0
3 years ago
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