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Xelga [282]
3 years ago
8

A'B'C is a translation of ABC. what is the length?​

Mathematics
2 answers:
kicyunya [14]3 years ago
6 0

Answer: 6 units

Step-by-step explanation:

andriy [413]3 years ago
3 0

Answer:

6 units

Step-by-step explanation:

A translation is a rigid body motion, meaning that it is a type of transformation that doesn't change the length of any sides, or it is the same as the original image. Since A'C' corresponds to AC, it must be 6 units as AC is 6 units.

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Simplify: log81/8 + 2log2/3 - 3log 3/2 +log 3/4​
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Answer:

0.

Step-by-step explanation:

Using the laws of logarithms:

log81/8 + 2log2/3 - 3log 3/2 + log 3/4​

= log 81/8 + log (2/3)^2 - log (3/2)^3 + log 3/4

= log 81/8 + log 4/9 - log 27/8 + log 3/4

= log 81/8 + log 4/9 - (log 27/8 - log 3/4)

= log (81/8 *  4/9) - log (27/8 * 4/3)

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7 0
4 years ago
1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co
Olenka [21]

Answer:

g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\

Step-by-step explanation:

First, notice that:

g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\

g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\

g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\

We proceed to use the chain rule to find g_{r}(\sqrt{2},\frac{\pi}{4}) using the fact that X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta) to find their derivatives:

g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\

Because we know X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta) then:

\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)

We substitute in what we had:

g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)

Now we put in the values r=\sqrt{2}\ and\ \theta=\frac{\pi}{4} in the formula:

g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})

Because of what we supposed:

g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})

And we operate to discover that:

g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}

g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}

and this will be our answer

3 0
3 years ago
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