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Masteriza [31]
3 years ago
6

Given QT = SR, QV = SU, and the diagram, prove that triangles QUT and SVR are congruent. Write a paragraph proof.

Mathematics
1 answer:
QveST [7]3 years ago
5 0

Answer:

Triangles QUT and SVR are congruent because the defining two sides and an included angle of triangles QUT and SVR are equal

Step-by-step explanation:

Here we have QT = SR and

QV = SU

Therefore,

QT = √(UT² + QU²)........(1)

RS = √(VS² + RV²)..........(2)

Since QS = QU + SU = QV + VS ∴ QU = VS

Therefore, since SR = QT and QU = VS, then from (1) and (2), we have UT = RV

Hence since we know all sides of the triangles QUT and SVR are equal and we know that the angle in between two congruent sides of the the triangles QUT and SVR that is the angle in between sides QU and UT for triangle QUT and the angle in between the sides RV and VS in triangle SVR are both equal to 90°, therefore triangles QUT and SVR are congruent.

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5+ 1/5x=5 what is x
Alenkasestr [34]

Answer:

x = 0

Step-by-step explanation:

Solve for x:

x/5 + 5 = 5

Put each term in x/5 + 5 over the common denominator 5: x/5 + 5 = x/5 + 25/5:

x/5 + 25/5 = 5

x/5 + 25/5 = (x + 25)/5:

(x + 25)/5 = 5

Multiply both sides of (x + 25)/5 = 5 by 5:

(5 (x + 25))/5 = 5×5

(5 (x + 25))/5 = 5/5×(x + 25) = x + 25:

x + 25 = 5×5

5×5 = 25:

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3 years ago
Read 2 more answers
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Step-by-step explanation:

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2 years ago
In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t 2 , 2 + t − 5t 2 , 1 + 2t} to the standard basis of P2.
Serjik [45]

Complete Question:

In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t² , 2+t− 5t² , 1 + 2t} to the standard basis C = {1, t, t²}. Then, write t² as a linear combination of the polynomials in B.

Answer:

The change of coordinate matrix is :

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

Step-by-step explanation:

Let U =  {D, E, F} be any vector with respect to Basis B

U = D [1 − 3t²] + E [2+t− 5t²] + F[1 + 2t]..............(*)

U = [D+2E+F]+ t[E+2F] + t²[-3D-5E]...................(**)

In Matrix form;

\left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right] \left[\begin{array}{ccc}D\\E\\F\end{array}\right] = \left[\begin{array}{ccc}D+2E+F\\E+2F\\-3D-5E\end{array}\right]

The change of coordinate matrix is therefore,

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

To find D, E, F in (**) such that U = t²

D + 2E + F = 0.................(1)

E + 2F = 0.........................(2)

-3D -5E = 1........................(3)

Substituting eqn (2) into eqn (1 )

D=3F...................................(4)

Substituting equations (2) and (4) into eqn (3)

-9F+10F=1

F = 1

Put the value of F into equations (2) and (4)

E = -2(1) = -2

D = 3(1) = 3

Substituting the values of D, E, and F into (*)

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

3 0
3 years ago
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