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3 years ago

How do you evaluate a function.

Mathematics

1 answer:

kumpel [21]3 years ago

3 0

Answer:

to evaluate a function, substitute the input

Step-by-step explanation:

(the given number or expression) for the function variable (place holder, x) replace the x with the number or expression. 1. given function f (x)=3x -5 finf (f) 4.

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Find the area of a triangle with a base length of 3 units and a height of 4 units.

svetlana [45]

Answer:

$6$ square units

Step-by-step explanation:

The area of a triangle can be calculated by multiplying the triangle's base by its height, and then dividing the result by $2$ . As an algebraic expression, this would be $\frac{bh}{2}$ , where $b$ is the triangle's base and $h$ is the triangle's height. In this case, we know that $b=3$ and $h=4$ . Therefore, the area of this triangle is $\frac{3*4}{2} =\frac{12}{2} =6$ square units. Hope this helps!

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2 years ago

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Lisa [10]

The answer is 2 ........

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3 years ago

Find the coordinates of the image of G(−3,9) after the translation (x,y)→(x+10,y−7) .

nataly862011 [7]

Answer:

D. (7,2)

Step-by-step explanation:

-3 +10 is 7

9-7 is 2.

kinda just put them together, you get (7, 2)

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2 years ago

What is the length of C'D' ?<br> feet

Alinara [238K]

Answer:

7 feet

Step-by-step explanation:

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3 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

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3 years ago