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3 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]3 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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BRAINLIEST TO THE BEST ANSWER !!!!

TiliK225 [7]

Which equation am I looking at

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3 years ago

The height 'h' (in feet) of a ball in a baseball game can be modeled by h = -16t² + 28t + 8 , where 't' is the time (in seconds)

lidiya [134]

Answer:

a) No. t < 0 is not part of the useful domain of the function

b) 2.0 seconds

Step-by-step explanation:

a) A graph of the function is shown below. It shows t-intercepts at t=-0.25 and t=2.0. We presume that t is measured forward from some event such as the ball being thrown or hit. The model's predicted ball location has no meaning prior to that event, when values of t are negative.

__

b) It is convenient to use a graphing calculator to find the t-intercepts. Or, the equation can be solved for h=0 any of several ways algebraically. One is by factoring.

h = 0 = -16t² +28t +8 . . . . . . . . . . . . the ball hits the ground when h = 0

0 = -4(4t² -7t -2) = -4(4t +1)(t -2)

This has t-intercepts where the factors are zero, at t=-1/4 and t=2.

The ball will hit the ground after 2 seconds.

4 0

3 years ago

Read these sentences from the text.

chubhunter [2.5K]

Answer: Hi! I would say it is D because the other answers really don’t make sense. And the passages said that they were rated low for air quality.

Step-by-step explanation:

I hope this helps!

3 0

3 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago

Help me with one and two plz

agasfer [191]

Answer:

1. parental function is g(x)=√x

2. parental function is g(x)=3√x

Step-by-step explanation:

1. To find the transformation, compare the function to the parent function and check to see if there is a horizontal or vertical shift, reflection about the x-axis or y-axis, and if there is a vertical stretch.

Parent Function: g(x)=√x

Horizontal Shift: None

Vertical Shift: Down 9 Units

Reflection about the x-axis: None

Vertical Stretch: Stretched

2. Parent Function: g(x)=3√x

Horizontal Shift: Left 5 Units

Vertical Shift: Up 3 Units

Reflection about the x-axis: Reflected

Vertical Compression: Compressed

6 0

3 years ago