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Romashka-Z-Leto [24]
3 years ago
6

How to find base area

Mathematics
2 answers:
ad-work [718]3 years ago
7 0
1) surface area = 18m^2 * pi * 26m = <span>8424pi m^2</span>
Tanzania [10]3 years ago
7 0
Pi multiplied by radius squared
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PLS HELP I DONT UNDERSTAND :( WILL GIVE BRAINLIEST!! +25 POINTS AHHHH
guajiro [1.7K]

Answer:

the anwser is A

Step-by-step explanation:

i understand

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4 years ago
Joe is 1.6 m tall. His shadow is 2 m long when he stands 3 m
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Answer:

Step-by-step explanation:

6 0
2 years ago
1) Solve for Side A.<br><br> 2) Solve for Angle B.
avanturin [10]

Answer:

solution given;

let

AB=a

AC=b=30ft

AB=c=20ft

<A=115°

By using Cosine rule.

a²=b²+c²-2bc cos angle

a²=30²+20²-2*30*20 Cos 115°

a²=1807.1419

a=√[1807.1419]

a=42.51

Side A is 42.51ft.

Again

Cos B=\frac{a²+c²-b²}{2ac}

Cos B=\frac{42.51²+20²-30²}{2*42.51*20}

Cos B=0.7687

<B=Cos -¹(0.7687)

<B=39.46°

Angle B is 39.46

4 0
3 years ago
Read 2 more answers
Let f be the function defined by f(x) = e^(x) cos x.
Pavel [41]
(a)

The average rate of change of f on the interval 0 ≤ x ≤ π is

   \displaystyle&#10;f_{avg\Delta} = \frac{f(\pi) - f(0)}{\pi - 0} =\frac{-e^\pi-1}{\pi}

____________

(b)

f(x) = e^{x} cos x \implies f'(x) = e^x \cos(x) - e^x \sin(x) \implies \\ \\&#10;f'\left(\frac{3\pi}{2} \right) = e^{3\pi/2} \cos(3\pi/2) - e^{3\pi/2} \sin(3\pi/2) \\ \\&#10;f'\left(\frac{3\pi}{2} \right) = 0 - e^{3\pi/2} (-1) = e^{3\pi/2}

The slope of the tangent line is e^{3\pi/2}.

____________

(c)

The absolute minimum value of f occurs at a critical point where f'(x) = 0 or at endpoints.

Solving f'(x) = 0

f'(x) = e^x \cos(x) - e^x \sin(x) \\ \\&#10;0 = e^x \big( \cos(x) - \sin(x)\big)

Use zero factor property to solve.

e^x \ \textgreater \  0\forall x \in \mathbb{R} so that factor will not generate solutions.
Set cos(x) - sin(x) = 0

\cos (x) - \sin (x) = 0 \\&#10;\cos(x) = \sin(x)

cos(x) = 0 when x = π/2, 3π/2, but x = π/2. 3π/2 are not solutions to the equation. Therefore, we are justified in dividing both sides by cos(x) to make tan(x):

\displaystyle\cos(x) = \sin(x) \implies 0 = \frac{\sin (x)}{\cos(x)} \implies 0 = \tan(x) \implies \\ \\&#10;x = \pi/4,\ 5\pi/4\ \forall\ x \in [0, 2\pi]

We check the values of f at the end points and these two critical numbers.

f(0) = e^1 \cos(0) = 1

\displaystyle f(\pi/4) = e^{\pi/4} \cos(\pi/4) = e^{\pi/4}  \frac{\sqrt{2}}{2}

\displaystyle f(5\pi/4) = e^{5\pi/4} \cos(5\pi/4) = e^{5\pi/4}  \frac{-\sqrt{2}}{2} = -e^{\pi/4}  \frac{\sqrt{2}}{2}

f(2\pi) = e^{2\pi} \cos(2\pi) = e^{2\pi}

There is only one negative number.
The absolute minimum value of f <span>on the interval 0 ≤ x ≤ 2π is
-e^{5\pi/4} \sqrt{2}/2

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(d)

The function f is a continuous function as it is a product of two continuous functions. Therefore, \lim_{x \to \pi/2} f(x) = f(\pi/2) = e^{\pi/2} \cos(\pi/2) = 0

g is a differentiable function; therefore, it is a continuous function, which tells us \lim_{x \to \pi/2} g(x) = g(\pi/2) = 0.

When we observe the limit  \displaystyle \lim_{x \to \pi/2} \frac{f(x)}{g(x)}, the numerator and denominator both approach zero. Thus we use L'Hospital's rule to evaluate the limit.

\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \lim_{x \to \pi/2} \frac{f'(x)}{g'(x)} = \frac{f'(\pi/2)}{g'(\pi/2)}

f'(\pi/2) = e^{\pi/2} \big( \cos(\pi/2) - \sin(\pi/2)\big) = -e^{\pi/2} \\ \\&#10;g'(\pi/2) = 2

thus

\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \frac{-e^{\pi/2}}{2}</span>

3 0
3 years ago
Solve for x. 39=3x simplify has much has you can
Sunny_sXe [5.5K]

Answer:

x=13

Step-by-step explanation:

Divide both sides by three to get rid of the three this makes x=13

6 0
3 years ago
Read 2 more answers
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