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Bezzdna [24]
2 years ago
10

Can someone please solve these problems for me if you could do step by step that would rlly help me this is due Feb 14, 2021 ple

ase HELP only do 7- 10 please ​

Mathematics
1 answer:
dsp732 years ago
7 0

Answer:

7. about 41.8 degrees or 0.72972 radians (used sin^{-1} (\frac{4}{6} ), because the given leg is opposite from the angle)

8. about 50.4 degrees or 0.87946 radians (used tan^{-1} (\frac{29}{24} ))

9. about 12.8 degrees or 0.224 radians (used sin^{-1} (\frac{12}{54} ))

10. about 51.3 degrees or 0.8957 radians (used cos^{-1} (\frac{25}{40} ), given leg is opposite from angle)

Step-by-step explanation:

S.O.H.--C.A.H.--T.O.A

sine = opposite (to angle)/hypotenuse

cosine = adjacent (to angle)/hypotenuse

tangent = opposite (to angle)/adjacent (to angle)

I didn't know if it was degrees or radians, so I did both! I saw you on the film question asking for help, so I just wanted to do so! Have a great day!

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In the diagram, what is mZVSR?
sesenic [268]

Answer:

80 degrees.

180-100=80

hope that helps

Please said thanks

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
Evaluate the following double integral where a = 2y
Keith_Richards [23]

Change the order of integration.

\displaystyle \int_0^1 \int_{2y}^2 \cos(x^2) \, dx \, dy = \int_0^2 \int_0^{x/2} \cos(x^2) \, dy \, dx \\\\ ~~~~~~~~ = \int_0^2 \cos(x^2) y \bigg|_{y=0}^{y=x/2} \, dx \\\\ ~~~~~~~~ = \frac12 \int_0^2 x \cos(x^2) \, dx

Substitute u=x^2 and du=2x\,dx.

\displaystyle \frac12 \int_0^2 x \cos(x^2) \, dx = \frac14 \int_0^4 \cos(u) \, du = \frac14 \sin(u) \bigg|_{u=0}^{u=4} = \boxed{\frac{\sin(4)}4}

4 0
1 year ago
In ∆ABC , the coordinates of the vertices are A(-2, 4), B(-1, 1), and C (2, 2). ∆ABC is a right triangle.
Akimi4 [234]

Answer: True

Step-by-step explanation: I made a picture and it adds up:

3 0
3 years ago
JK, KL, and JL are all tangent to circle O. JA = 3, JL = 7, and the perimeter of triangle JKL = 26. What Are JK and KL?
sattari [20]
Answer is B) JK=9 and KL=10

you subtract 7 from 26 to get 19 

4 0
3 years ago
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