Find the length of a hypotenuse whose legs are 5 and 6

What is the range of possible values for x?

Talja [164]

The minimum value for 2x is 0
the maximum value is achieved when A, D and C are collinear and the quadrilateral ABCD becomes an isosceles triangle ABC 
base AB = 52 and vertical angle 2x + 34° 

For the sine law 
(sin 2x)/22 = (sin ADB)/AB 
(sin 34°)/30 = (sin BDC)/BC 

is given that AB = BC, and sin ADC = sin BDC because they are supplementary, so from 
(sin ADC)/AB = (sin BDC)/BC 

it follows 
(sin 2x)/22 = (sin 34°)/30 

sin 2x = 22 (sin 34°)/30 

2x = asin(22 (sin 34°)/30) ≈ 24.2° 

x = 0.5 asin(22 (sin 34°)/30) ≈ 12.1° 

0 < x < 12.1°

7 0

3 years ago

In winter, the price of apples suddenly went up by$0.75 per pound. Sam bought 3 pounds of apples at the new price for a total of

White raven [17]

From the information given, the original price per pound was increased by $ 0.75
Let the original price be x, so the new price is x + 0.75

Sam bought 3 pounds of apples at new price and it cost him 5.88

so we form an equation:

3 (x + 0.75) = 5.88

3x + 2.25 = 5.88

3x = 5.88 - 2.25

3x = 3.63

x = 3.63 / 3

x = 1.21

Therefore the original price per pound is 1.21 dollars

4 0

3 years ago

Read 2 more answers

Triangle PQR is rotated 180° clockwise about point M. The image of point P is P' and the image of point R is R'. What are the co

Mama L [17]

Answer: C= (1,2)

Step-by-step explanation:

7 0

3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago

Jackrabbits are capable of reaching speeds up to 40 miles per hour. How fast is this in feet per second? (Round to the nearest w

Contact [7]

There are 5280 feet in a mile
There are 3600 seconds in an hour

to get 40 miles per hour
we multiply 40 by 5280 and divide by 3600

40 * 5280 / 3600

this gets us 58.666666...
but since they want us to round to the nearest whole number, our answer would be 59 feet per second

5 0

3 years ago