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Olenka [21]
3 years ago
10

Question 17 PLEASE HELP given the function: f(x) =3x+7. Find the value of y when x=-8

Mathematics
1 answer:
STALIN [3.7K]3 years ago
8 0

→Answer:

y = -17

→Step-by-step explanation:

Well first off the f(x) in the equation f(x) = 3x + 7 can also be written as y,

y = 3x + 7

So if x is -8 we need to plug it in for x and solve, simple as that :)

y = 3(-8) + 7

y = -24 +7

y = -17

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Write each product using an exponent.<br><br> 9 × 9 × 9 × 9
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9^4 is what you’re looking for right?

Step-by-step explanation:

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A bouncy ball is dropped such that the height of its first bounce is 2.75 feet and each successive bounce is 80% of the previous
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7 0
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

3 0
3 years ago
NEED HELP FAST PLEASE!!!
trasher [3.6K]
Answer: Choice B

The slope is rise/run = 2/3 meaning we go up 2 and over to the right 3. For instance, moving from (0,0) to (3,2) has us move up 2 and over to the right 3.

Alternatively you can use the slope formula m = (y2-y1)/(x2-x1) to get the same result of 2/3.
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