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Simora [160]
3 years ago
13

Help please! the picture is below!

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
4 0
The answer would be B
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I hope this helps you

3 0
3 years ago
If two figures are congruent, their area are equal. the area if ABCD equals the area if PQRS​
JulsSmile [24]

Answer:

Yes

Step-by-step explanation:

Since the sides are each congruent then ABCD and PQRS are the exact same shape and size. This means the area inside each figure will be the same amount. The areas are the same.

7 0
3 years ago
The 50th term of an arithmetic sequence is 86 and the common difference is 2.  Find the first three terms of the sequence.
Nesterboy [21]
a_n=a_{n-1}+d=a_{n-2}+2d=\cdots=a_1+(n-1)d

which means

a_{50}=86=a_1+49\times2\implies a_1=-12

which in turn means the next two terms are a_2=-10 and a_3=-8.
6 0
3 years ago
!!!10 POINTS I KNOW IT IS NOT ENOUGH BUT PLEASE ANSWER!
mart [117]

Answer:

B) 376 units squared

Step-by-step explanation:

Imma put a picture that has a fully explanation. Hope this helps, good luck! ;)

Btw saw your other one got deleted •_• *sigh*

4 0
3 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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