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2 years ago

This diagram shows triangles ABC and PQR.

Mathematics

1 answer:

Stels [109]2 years ago

6 0

Answer:

<h3>A reflection across the line x=3, a reflection across the x-axis and a dilation with a scale factor of 2, because each side is double.</h3><h3>Step-by-step explanation:</h3>

We know that the first transfomration is a rotation 90° clockwise.

Notice that vertex R is at the same horizontal coordinate than vertex C, which means the second transformation must include a reflection across the line x=3, a reflection across the x-axis and a dilation with a scale factor of 2, because each side is double.

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Step-by-step explanation:

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2 years ago

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3 years ago

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Dafna11 [192]

$\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6$

therefore
$D:x\in\left< -\dfrac{5}{4};\ 6\right>$

$\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs$
$9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D$
Answer: t = -1.

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2 years ago

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