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dlinn [17]
3 years ago
15

In ∆PQR, ∠P = 90degrees , PQ= 12 cm, PR=16cm. Find ∠Q

Mathematics
1 answer:
siniylev [52]3 years ago
5 0

Answer:

∠Q  = 53.13 degrees

Step-by-step explanation:

Given that ∆PQR, ∠P = 90degrees , this means that the triangle is a right angled triangle

Hence using the notation

SOH CAH TOA

where S is sine, C is cosine, T is tangent and O, A and H represents the size of the opposite, adjacent and hypotenuse sides

Considering  ∠Q and the given sides

PR=16cm is the opposite side,

PQ= 12 cm is the adjacent side hence we use TOA

Tan Q = 16/12 =

Q = Arc tan 16/12

= 53.13 degrees

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The answer is D. Note that the y-intercept is -3, not 3. So, when x is at 0, y should be -3. Choice D is the only graph with a y-intercept of -3.
7 0
3 years ago
5+(x-5)=x justify each step
max2010maxim [7]

5+(x-5)=x

Simplify the left side by combining like terms:

5 + x - 5 = x

5-5 = 0

x = x

X = All real numbers.

4 0
3 years ago
Read 2 more answers
A steamer travels 36 km upstream and 32 km downstream in 6.5 hours. The same steamer travels 4 km upstream and 40 km downstream
rewona [7]

Answer:

The streamer's speed in still water is 90.23 km/h while the stream's speed is 33.33 km/h

Step-by-step explanation:

Let v = streamer's speed in still water and v' = stream's speed. His speed upstream is V = v + v' and his speed downstream is V' = v - v'.

Since he travels 36 km upstream, the time taken is t = 36/V = 36/(v + v').

he travels 32 km downstream, the time taken is t' = 32/V' = 32/(v - v')

The total time is thus t + t' = 36/(v + v') + 32/(v - v')

Since the whole trip takes 6,5 hours,

36/(v + v') + 32/(v - v') = 6.5  (1)

Multiplying each term by (v + v')(v - v'), we have

(v + v')(v - v')36/(v + v') + (v + v')(v - v')32/(v - v') = 6.5(v + v')(v - v')  (1)

(v - v')36 + (v + v')32 = 6.5(v + v')(v - v')  (1)

36v - 36v' + 32v + 32v' = 6.5(v² + v'²)

68v - 4v' = 6.5(v² + v'²)        (2)

Also he travels 4 km upstream, the time taken is t" = 4/V = 4/(v + v').

he travels 40 km downstream, the time taken is t'" = 40/V' = 40/(v - v')

The total time is thus t" + t'" = 4/(v + v') + 40/(v - v')

Since the whole trip takes 180 minutes = 3 hours,

4/(v + v') + 40/(v - v') = 3  (3)

Multiplying each term by (v + v')(v - v'), we have

(v + v')(v - v')4/(v + v') + (v + v')(v - v')40/(v - v') = 3(v + v')(v - v')  (1)

(v - v')4 + (v + v')40 = 3(v + v')(v - v')  (1)

4v - 4v' + 40v + 40v' = 3(v² + v'²)

44v - 36v' = 3(v² + v'²)      (4)

Dividing (2) by (4), we have

(68v - 4v')/(44v - 36v') = 6.5(v² + v'²)/3(v² + v'²)      

(68v - 4v')/(44v - 36v') = 6.5/3

3(68v - 4v') = 6.5(44v - 36v')

204v - 12v' = 286v - 234v'

204v - 286v = 12v' - 234v'

-82v = -222v'

v = -222v'/82

v = 111v'/41

Substituting v into (2), we have

68v - 4v' = 6.5(v² + v'²)      

68(111v'/41) - 4v' = 6.5[(111v'/41)² + v'²]        

[68(111/41) - 4]v' = 6.5[(111/41)² + 1]v'²    

[68(111/41) - 4]v' = 6.5[(111/41)² + 1]v'²

[7548/41 - 4]v' = 6.5[12321/1681 + 1]v'²

[(7548 - 164)/41]v' = 6.5[(12321 + 1681)/1681]v'²

[7384/41]v' = 6.5[14002/1681]v'²

[7384/41]v' = [91013/1681]v'²

[91013/1681]v'² - [7384/41]v' = 0

([91013/1681]v' - [7384/41])v' = 0

⇒ v' = 0 or ([91013/1681]v' - 7384/41) = 0

⇒ v' = 0 or [91013/1681]v' - 7384/41) = 0

⇒ v' = 0 or v' =  7384/41 × 16841/91013

⇒ v' = 0 or v' =  180.097 × 0.185

⇒ v' = 0 or v' =  33.33 km/h

Since v' ≠ 0, v' = 33.33 km/h

Substituting v' into v = 111v'/41 = 111(33.33 km/h)/41 = 3699.63 km/h ÷ 41 = 90.23 km/h

So, the streamer's speed in still water is 90.23 km/h while the stream's speed is 33.33 km/h

5 0
3 years ago
let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a
tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b)  two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) =  y = 0(x) + 1(y)

Matrix Representation of T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]

now, eigen value of 'T'

T - kI =  \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b)  two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

#SPJ4

6 0
1 year ago
Item 32 On three 150-point geography tests, you earned grades of 88%, 94%, and 90%. The final test is worth 250 points. What per
stira [4]
97.2% on the final.
Find the total amount of points needed to get a 93%. 150+150+150+250 = 700 total points. 700*.93 = 651 total points needed to get 93%. find the points received from the other quizzes and subtract the sum from 651 to find how many points need to be scored on the final.
150*.88=132 (test 1)
150*.94=141 (test 2)
150*.90=135 (test 3)
132+141+135= 408

651-408 = 243 points to score on the final exam.
243/250 =.972 or 97.2%
3 0
3 years ago
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