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3 years ago

Use the diagram and given information to answer the questions and prove the statement.

Mathematics

1 answer:

kiruha [24]3 years ago

4 0

Answer:

See explanation

Step-by-step explanation:

Consider triangles BXY and AZY. In these triangles:

$\angle X\cong \angle Z$ - given;
$\overline{XY}\cong \overline{ZY}$ - given;
$\angle XYZ\cong \angle ZYA$ as right angles by reflective property.

Hence, $\triangle BXY\cong \triangle AZY$ by ASA postulate (or by LA postulate).

Congruent triangles have congruent corresponding sides, so

$\overline{XB}\cong \overline{ZA}$

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alexgriva [62]

Answer:

You can put this solution

30.24 - 2.24 = $28 just tickets

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(30.24 - 2.24)/x

Step-by-step explanation:

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2 years ago

10. When Raphael solved the system below by using the method of elimination, he changed the second equationto -8x – 2y = -2. Wha

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The property that justifies multiplying through eqn(2) by -2 is;

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1 year ago

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Step-by-step explanation:

In given rt.angled triangle,

Sin thita =p/h

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2 years ago

Evaluate the expression.<br> 3–6–(2+6)2

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3 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago