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3 years ago

29

Mathematics

1 answer:

kramer3 years ago

4 0

Answer: C (1.25, 10.25)

Step-by-step explanation:

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adoni [48]

Step-by-step explanation:

CCTV cameras in a bit more of the day I will have a great time and where is the only thing that has to do with it

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3 years ago

39. State the equation for the line described below.

devlian [24]

Answer:

Step-by-step explanation:

39). y = $-\frac{7}{8}$ x + 3

40). m = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

y - $y_{1}$ = m( x - $x_{1}$ )

(- 7, 5)

( - 12, 4)

m = ( 4 - 5 ) / ( - 12 + 7) = 1/5

y - 5 = (1/5)(x + 7) ⇒ y = $\frac{1}{5}$ x + 6.4

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2 years ago

Hans leaned a 17 ft ladder against his house. He placed the base of the ladder 8 ft from his house, and the top of the ladder re

Fynjy0 [20]

Answer:

Step-by-step explanation:

so 17 squared minus 8 squared is 225. the square root of 225 is 15

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3 years ago

The number of tickets collected by joshua,kenneth,and larry is in the ratio 2:5:8.kenneth collected 85 tickets

mart [117]

I don't understand you questions

3 0

3 years ago

Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

7 0

3 years ago