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Fofino [41]
3 years ago
11

Find the map ratio if 10 cm on the map is represented by 10 km

Mathematics
2 answers:
olga nikolaevna [1]3 years ago
4 0
I think this is the answer :)

Dafna11 [192]3 years ago
4 0

Answer:

Step-by-step explanation:

1 km = 100000cm

10 km = 10*100000 = 1,000,000cm

10cm : 10 km = 10 : 1,000,000 = 1: 100,000

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Well 1.75ft times 1000 is 1750ft of cable
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Becky spends 3/5 of her money she has left £60 left. how much money did she start with?
Virty [35]

Answer: $150

Step-by-step explanation:

Becky spends 3/5 of her Money and is left with 2/5 of her money.

Hence 2 parts = $60

And 5 parts = $150

Becky has $150 at first.

5 0
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Please help me post test !!!!!!
Olin [163]

Answer:

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Step-by-step explanation:

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5 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
How do you write 3.684 × 103 in standard form?
lys-0071 [83]
The answer is 379.452
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3 years ago
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