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2 years ago

10

in the adjoining figure P A and PB are tangents from P to a circle with Centre O if angle APB is equal to 40 degree then find an

gle ACB

1 answer:

Anna11 [10]2 years ago

5 0

Step-by-step explanation:

40° + 90° + 90° + angle ACB = 360°

220° + angle ACB = 360°

angle ACB = 360° - 220°

Therefore angle ACB = 140°

Hope it will help :)

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Which of the following values for m proves that 2m + 2m is not equivalent to 4m2?

Mashutka [201]

M=0 i am not sure it is right

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2 years ago

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A brochure from the department of public safety in a northern state recommends that motorists should carry 12 items (flashlights

scoray [572]

Answer:

$Mean = 6.07$

$Median = 7$

$Mode = 7$

Step-by-step explanation:

Given

$Data: 5\ 3\ 7\ 8\ 0\ 1\ 0\ 5\ 12\ 10\ 7\ 6\ 7\ 11\ 9$

$n = 15$

Solving (a): The mean

Mean is calculated as:

$Mean = \frac{\sum x}{n}$

This gives:

$Mean = \frac{5+ 3+ 7+ 8+ 0+ 1+ 0+ 5+ 12+ 10+ 7+ 6+ 7+ 11+ 9}{15}$

$Mean = \frac{91}{15}$

$Mean = 6.07$

Solving (b): The median

Sort the data in ascending order:

$Data: 5\ 3\ 7\ 8\ 0\ 1\ 0\ 5\ 12\ 10\ 7\ 6\ 7\ 11\ 9$

$Sorted: 0\ 0\ 1\ 3\ 5\ 5\ 6\ 7\ 7\ 7\ 8\ 9\ 10\ 11\ 12$

The median is:

$Median = \frac{n + 1}{2}th$

$Median = \frac{15 + 1}{2}th$

$Median = \frac{16}{2}th$

$Median = 8th$

The 8th item on the sorted dataset is 7; So:

$Median = 7$

Solving (c): The mode

$Mode = 7$

Because it has a frequency of 3 (more than any other element of the dataset).

6 0

2 years ago

A ferry will safely accommodate 68 tons of passenger cars. Assume that the mean weight of a passenger car is 1.8 tons with stand

nika2105 [10]

Answer:

$P (Z$

Step-by-step explanation:

We want the probability that the 35 cars are loaded onto the ferry. Therefore:

Since

μ=1.8 and σ=0.5 we have:

P(X<35 )=P ( X−μ<35−1.8 )=P((X−μ)/σ<(35−1.8)/0.5)

Since

(x−μ)/σ=Z and

$(35-1.8)/0.5=66.4$

we have:

$P (X$

Use the standard normal table to conclude that:

$P (Z$

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2 years ago

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Three trigonometric functions for a given angle are shown below. cosecant theta = thirteen-twelfths; secant theta = Negative thi

Kay [80]

Answer:

(–5, 12) is the correct answer.

Step-by-step explanation:

We are given the following values:

$cosec\theta = \dfrac{13}{12}\\sec\theta=-\dfrac{13}{5}\\cot\theta =-\dfrac{5}{12}$

Now, we know the following identities:

$sin \theta = \dfrac{1}{cosec\theta}\\cos \theta = \dfrac{1}{sec\theta}\\tan \theta = \dfrac{1}{cot\theta}$

Now, the values are:

$sin\theta = \dfrac{12}{13}\\cos\theta=-\dfrac{5}{13}\\tan\theta =-\dfrac{12}{5}$

Sine value is positive and cos, tan values are negative.

It can be clearly observed that $\theta$ is in 2nd quadrant.

2nd quadrant means, the value of $x$ will be negative and $y$ will be positive.

Let us have a look at the value of $tan\theta$ :

$tan\theta = \dfrac{Perpendicular}{Base}\\OR\\tan\theta = \dfrac{y-coordinate}{x-coordinate} = -\dfrac{12}{5}\\\therefore y = 12,\\x = -5$

Please refer to the attached image for clear understanding and detailed explanation.

Hence, the correct answer is coordinate (x,y) is (–5, 12)

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2 years ago

Coordinates, gradient and tangent work (see image).

Ivenika [448]

Answer:

a) $P(x,2x^2-5),\ Q(x+h,2(x+h)^2-5)$

b) $4x+2h$

c) $4x$

Step-by-step explanation:

Given the curve

$y=2x^2-5$

a) If the x-coordinate of P is $x$ , then the y-coordinate is $2x^2-5,$ so point P has coordinates $(x,2x^2-5)$

If the x-coordinate of Q is $x+h$ , then the y-coordinate is $2(x+h)^2-5$ so point Q has coordinates $(x+h,2(x+h)^2-5)$

b) The gradient of the secant RQ is

$\dfrac{y_Q-y_P}{x_Q-x_P}\\ \\=\dfrac{(2(x+h)^2-5)-(2x^2-5)}{(x+h)-x}\\ \\=\dfrac{2(x+h)^2-5-x^2+5}{x+h-x}\\ \\=\dfrac{2(x+h)^2-2x^2}{h}\\ \\=\dfrac{2x^2+4xh+2h^2-2x^2}{h}\\ \\=\dfrac{4xh+2h^2}{h}\\ \\=4x+2h$

c) If $h\rightarrow 0,$ then the gradient $4x+2h\rightarrow 4x$

5 0

2 years ago