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2 years ago

Find the common ratio r for the geometric sequence and use r to find the next three terms.

Mathematics

2 answers:

Lemur [1.5K]2 years ago

8 0

Answer:

r=3 three terms: 3240, 9720, and 29160

Step-by-step explanation:

120/40=3
1080*3=3240*3=9720*3=29160

lesantik [10]2 years ago

7 0

Answer:

3240, 9720, 29160

Step-by-step explanation:

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∠A and ∠B are vertical angles. ∠A = 65x − 12 and ∠B = 43x + 10 How many degrees are in ∠A?

Margarita [4]

Answer:

53 degrees

Step-by-step explanation:

Vertical angles are congruent so...

65x-12=43x+10

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Then add it into the equation of ∠A

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2 years ago

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Answer:

Step-by-step explanation:

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2 years ago

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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

2 years ago

What are other ways to write 20,000,000

Klio2033 [76]

2×10 to the 7th power

3 0

3 years ago