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3 years ago

10

Please answer I will give you brainliest!!! Find the angles of the rhombus if the ratio of the angles formed by diagonals and th

e sides of the rhombus is 6:5

1 answer:

Aleksandr-060686 [28]3 years ago

7 0

Answer:

one angle : 6x=6(90/11)=540/11 (49.1 rounded to the nearest tenth)

other angle=5(90/11)=450/11 (41 degrees rounded to the nearest ones)

Step-by-step explanation:

ratio of the angles 6:5

when you draw diagonal inside a rhombus it creates four right angled triangle

one angle=90

6x+5x=90 ( the other angles is complimentary)

11x=90

x=90/11

one angle : 6x=6(90/11)=540/11 (49.1 rounded to the nearest tenth)

other angle=5(90/11)=450/11 (41 degrees rounded to the nearest ones)

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3 years ago

What are the coordinates of the x-intercepts of the parabola y = x² - 8x + 15?

tamaranim1 [39]

Answer:

(3, 0) and (5, 0)

Step-by-step explanation:

we have

$y=x^{2}-8x+15$

we know that

The x-intercepts are the values of x when the value of y is equal to zero

so

For y=0

$x^{2}-8x+15=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

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in this problem we have

$x^{2}-8x+15=0$

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substitute in the formula

$x=\frac{-(-8)\pm\sqrt{-8^{2}-4(1)(15)}} {2(1)}$

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4 0

3 years ago

You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then

gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is $d\theta_1 =(14.36p)^o$

Step-by-step explanation:

From the question we are told that

The angle of elevation is $\theta_1 = 15 ^o = \frac{\pi}{12}$

The height of the tree is h

The distance from the base is D

h is mathematically represented as

$h = D tan \theta$ Note : this evaluated using SOHCAHTOA i,e

$tan\theta = \frac{h}{D}$

Generally for small angles the series approximation of $tan \theta \ is$

$tan \theta = \theta + \frac{\theta ^3 }{3}$

So given that $\theta = 15 \ which \ is \ small$

$h = D (\theta + \frac{\theta^3}{3} )$

$dh = D (1 + \theta^2) d\theta$

=> $\frac{dh}{h} = \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta$

Now from the question the relative error of height should be at most

$\pm p$ %

=> $\frac{dh}{h} = \pm p$

=> $\frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta = \pm p$

=> $d\theta = \pm \frac{\theta + \frac{\theta^3}{3} }{1+ \theta ^2} * \ p$

So for $\theta_1$

$d\theta_1 = \pm \frac{\theta_1 + \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} * \ p$

substituting values

$d [\frac{\pi}{12} ] = \pm \frac{[\frac{\pi}{12} ] + \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} * \ p$

=> $d\theta_1 = 0.25 p$

Converting to degree

$d\theta_1 = (0.25* 57.29) p$

$d\theta_1 =(14.36p)^o$

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3 years ago

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Sladkaya [172]

You are correct

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3 years ago

Read 2 more answers