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andriy [413]
3 years ago
12

Which is the hypotenuse-angle theorem? If the hypotenuse and an acute angle of a right triangle are congruent to the correspondi

ng parts of another right triangle, then the triangles are congruent. If the hypotenuse and an obtuse angle of a right triangle are congruent to the corresponding parts of another right triangle, then the triangles are congruent. If the hypotenuse and an acute angle of a right triangle are congruent to the corresponding parts of another right triangle, then the triangles are complimentary. If the hypotenuse and an acute angle of a right triangle are congruent to the corresponding parts of another right triangle, then the triangles are supplementary.
Mathematics
1 answer:
Rufina [12.5K]3 years ago
8 0
i would say this is the third one
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What is 7+3t over 4 = -t over 8
katrin [286]
7+3t       -t
------- = -----  is the relevant equation here.  We don't ask, "What is 7+3t over
   4          8    -t over 8," but rather "Solve the following equation for t."

Cross multiplying, 56 + 24t = -4t
 
                                       -56
Then:  28t = 56, and t = ------ = -2 (answer)
                                         28

The solution to "<span>What is 7+3t over 4 = -t over 8" is t = -2.</span>
5 0
4 years ago
A store sells 4 cans of nuts for $7. How much would it cost you to buy 7 cans of nuts?
maksim [4K]

Answer:

12.25

i think thats right

cause you would divided the 7 by four to get you 1.75 then multiply it by 7

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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
Translated 2 units right
melisa1 [442]

Answer:

yes

Step-by-step explanation:

3 0
3 years ago
25
Elodia [21]

Answer:

33 degrees

Step-by-step explanation:

(4x+1) and (2x-19) add up to 180 degrees

(4x+1)+(2x-19)=180\\6x-18=180\\6x=198\\x=33

3 0
2 years ago
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