What is the mapping notation that will transform triangle ABC to triangle A'B'C'?

Mathematics

1 answer:

MissTica3 years ago

6 0

Answer:

I cant see it so can u do something so i can see it plz

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Question 15 (5 points)<br> Find the angle between u = <7, -2> and v= (-1,2>.

Tema [17]

Answer:

Approximately $2.3127$ radians, which is approximately $132.51^{\circ}$ .

Step-by-step explanation:

Dot product between the two vectors:

$\begin{aligned}& u\cdot v \\ =\; & 7 \times (-1) + (-2) \times 2 \\ =\; & (-11) \end{aligned}$ .

Magnitude of the two vectors:

$\begin{aligned} \| u \| &= \sqrt{{7}^{2} + {(-2)}^{2}} \\ &= \sqrt{53} \end{aligned}$ .

$\begin{aligned} \| v \| &= \sqrt{{(-1)}^{2} + {2}^{2}} \\ &= \sqrt{5} \end{aligned}$ .

Let $\theta$ denote the angle between these two vectors. By the property of dot products:

$\begin{aligned} \cos(\theta) &= \frac{u \cdot v}{\|u\| \, \| v \|} \\ &= \frac{(-11)}{(\sqrt{53})\, (\sqrt{5})} \\ &= \frac{(-11)}{\sqrt{265}}\end{aligned}$ .

Apply the inverse cosine function ${\rm arccos}$ to find the value of this angle:

$\begin{aligned} \theta &= \arccos\left(\frac{u \cdot v}{\| u \| \, \| v \|}\right) \\ &= \arccos\left(\frac{(-11)}{\sqrt{265}}\right) \\ & \approx \text{$2.3127$ radians} \\ &= 2.3127 \times \frac{180^{\circ}}{\pi} \\ &\approx 132.51^{\circ}\end{aligned}$ .