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JulijaS [17]
2 years ago
6

What are the values of x and y?

Mathematics
1 answer:
Vesna [10]2 years ago
3 0

Answer:

B)   x = 8\sqrt{3} ;  y = 4\sqrt{3}

Step-by-step explanation:

in a 30-60-90° triangle the sides, respectively, are in the ratio of

1 : \sqrt{3} : 2

we can find 'x' by creating this proportion:

12/\sqrt{3} = x/2

cross-multiply to get:

\sqrt{3}x = 24

x = 24/\sqrt{3}

we can simplify this by 'rationalizing the denominator' which is to

multiply numerator and denominator by \sqrt{3} to get:

(24\sqrt{3})÷3, which is 8\sqrt{3}

we know that side 'y' is half that of side 'x'

therefore y = 4\sqrt{3}

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Travka [436]
The given histogram represents <span>the number of hamburgers students ate in a month :
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From the histogram we can conclude the following:
(1) 8 students ate (0 - 4) hamburgers 
(2) 3 students ate (5 - 9) <span>hamburgers </span>
(3) 2 students ate (10 - 14) <span>hamburgers
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note: we don't know how many students ate exactly 5 hamburgers.

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7 0
2 years ago
Read 2 more answers
I need help with this problem
Arada [10]

Answer:

65.56°

Step-by-step explanation:

We know that if we take dot product of two vectors then it is equal to the product of magnitudes of the vectors and cosine of the angle between them

That is let p and q be any two vectors and A be the angle between them

So, p·q=|p|*|q|*cosA

⇒cosA=\frac{u.v}{|u||v|}

Given u=-8i-3j and v=-8i+8j

|u|=\sqrt{(-8)^{2}+ (-3)^{2}} =8.544

|v|=\sqrt{(-8)^{2}+ (8)^{2}} =11.3137

let A be angle before u and v

therefore, cosA=\frac{u.v}{|u||v|}=\frac{(-8)*(-8)+(-3)*(8)}{8.544*11.3137} =\frac{40}{96.664}

⇒A=arccos(\frac{40}{96.664} )=arccos(0.4138 )=65.56

Therefore angle between u and v is 65.56°

5 0
2 years ago
X^2+2x-8=0 in simplest radical form
madreJ [45]

2⃣, -4⃣ = x; no radical necessary.

6 0
2 years ago
Section 5.2 Problem 6:<br><br>Find the general solution<br><img src="https://tex.z-dn.net/?f=y%27%27%20%2B%206y%27%20%2B%2010y%2
mihalych1998 [28]

Answer:

y=e^{-3t}(A\: cos\: t+B\:sin\:t)

Step-by-step explanation:

<u>Given Second-Order Homogenous Differential Equation</u>

y''+6y'+10y=0

<u>Use Auxiliary Equation</u>

<u />m^2+6m+10=0\\\\(m+3)^2+1=0\\\\(m+3)^2=-1\\\\m+3=\pm i\\\\m=-3\pm i

<u>General Solution</u>

<u />y=e^{pt}(A\: cos\: qt+B\:sin\:qt)\\\\y=e^{-3t}(A\: cos\: t+B\:sin\:t)

Note that the DE has two distinct complex solutions p\pm qi where A and B are arbitrary constants.

4 0
2 years ago
Help im stuck on this problem
kotykmax [81]

Answer:

J.

{ \bf{m \angle A}} = { \bf{180 \degree -48 \degree }} \\ { \tt{m \angle A = 132  \degree}}

6 0
3 years ago
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