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Alex787 [66]
3 years ago
14

Which of the following best describes triangle TUV and triangle XYZ? Please help!

Mathematics
1 answer:
Vesna [10]3 years ago
7 0

Answer:

Option C. Triangle TUV is similar to triangle XYZ

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

If two figures are congruent, then its corresponding sides and its corresponding angles are congruent

In this problem, the corresponding angles between triangle TUV and triangle XYZ are congruent, but its corresponding sides are not congruent

therefore

The triangles are similar

so

The ratio of its corresponding sides must be proportional

<em><u>Verify</u></em>

\frac{XY}{TU}=\frac{YZ}{UV}

substitute the values

\frac{6}{3}=\frac{6}{3V}

2=2 ---> is ok

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= (-4,5)

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Step-by-step explanation:

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Step-by-step explanation:

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If x+ay =b and<br> ax-by=c <br> Then find out x , y . (with process)
n200080 [17]

Answer:

The solution of the given system of equations is,

x=\frac{b^2+ac}{a^2+b};\; y=\frac{ab-c}{a^2+b}

Step-by-step explanation:

The given equations are :

<em>x</em> + a<em>y</em> = b          .......(1)

a<em>x</em> - b<em>y</em> = c        .......(2)

We will use 'Substitution Method' to solve the given system of equations.

In this method, we will find out the value of either of the two variables that is '<em>x</em>' and '<em>y</em>' from one of the two equations in terms of the another variable  and then substitute that value in the other equation to find the value of the another variable.

Now, we will be finding out the value of '<em>x</em>' in terms of '<em>y</em>' from equation (1) and then substitute it in the equation (2).

Consider the equation (1), that is,

   <em>x</em> + a<em>y</em> = b ⇒ <em>x</em> = b - a<em>y</em>          ........(3)

Substitute the value of '<em>x</em>' from (3) in (2), we get

  a(b - a<em>y</em>) - b<em>y</em> = c

⇒ab - a²<em>y</em> - b<em>y</em> = c

⇒-a²<em>y</em> - b<em>y</em> = c - ab

⇒-<em>y</em>(a²+b) = c - ab

⇒<em>y</em>(a²+b) = ab - c

\implies y=\frac{ab-c}{a^{2}+b}

Now, substituting the above value of '<em>y</em>' in equation (3), we get

x=b-a(\frac{ab-c}{a^2+b})

\implies x=\frac{b(a^2+b)-a(ab-c)}{a^2+b}

\implies x=\frac{a^2b+b^2-a^2b+ac}{a^2+b}

\implies x = \frac{b^2+ac}{a^2+b}

Hence, the solution of the given system of equations is,

x = \frac{b^2+ac}{a^2+b} ;\; y = \frac{ab-c}{a^2+b}

5 0
3 years ago
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