$11.20 is 30% of what number

Firdavs [7]

Look at the link below I filled in the answers for you. :)

6 0

3 years ago

The length of an equilateral triangle is increased by 5 cm. The perimeter of the new triangle is 72 cm. What is the side length

Igoryamba

Answer:

19cm

Step-by-step explanation:

An equilateral triangle is a triangle where the length of the 3 sides are the same.

the perimeter is the sum of the 3 sides

for example, if the length of the side of the an equilateral triangle is 5, the perimeter is 5 x 3 = 15

let x = unknown length of side

length after increase = x + 5

(x + 5) x 3 = 72

divide both sides of the equation by 3

x + 5 = 24

subtract 5 from both sides of the equation

x = 24 - 5 = 19cm

6 0

3 years ago

10. Does the function represent a linear or nonlinear function?<br> For 10 or 11

solniwko [45]

I believe their both linear because their in the correct form

4 0

3 years ago

Read 2 more answers

Express Jenny's age in terms of a if jenny is three years less than half of Anne's age

Nadusha1986 [10]

1. a = 1/2x - 3
2. a = 2x + 7

3 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