Erika invested $6,000 in a bond at a yearly rate of 3% for 2.5 years. How much interest did she earn?

What is the equation of this graphed line?

liq [111]

You can put the line into the equation y= MX+c
m = gradient
c = y offset

To work out m you need to divide the change in y by the change in x for two points on the graph:
6--6 = 12 (change in x)
-3--7 = 4 (change in y)

m = 4/12 = 1/3

Then to work out the c variable you need to see where the line crosses the y axis, where this happens is the c variable:
c = -5 because c crosses y at the point -5

Then place it into one equation:
y = 1/3x-5

8 0

3 years ago

A local Charity receives 1/3 of funds raised during a craft fair and bake sale. The total amount given to the charity was $137.4

Fed [463]

Answer:

The bake sale raised $412.35

Step-by-step explanation:

* Lets explain how to solve the problem

- A local Charity receives 1/3 of funds raised during a craft fair and

bake sale

∴ The total amount given to the charity = 1/3 × the funds raised during

a craft fair and bake sale

- The total amount given to the charity was $137.45

∴ The 1/3 of of funds raised during a craft fair and bake sale is $137.45

- Lets substitute the The total amount given to the charity in the

equation above to find the funds raised during the bake sale

∴ 137.45 = 1/3 × the funds raised during the bake sale

- Multiply the both sides by 3

∴ 3 × 137.45 = the funds raised during the bake sale

∴ The funds raised during the bake sale = 412.35

* The bake sale raised $412.35

6 0

3 years ago

What is the value of x? Enter your answer in the box. |x = °|

Serga [27]

X+30+70=180
x+100=180. -100 both sides
x=80

|x=80° |

8 0

4 years ago

Is the following relation a function? ху 1 -2 1-3 2 1 3 - 2 Yes No

professor190 [17]

Answer:

no

Step-by-step explanation:

In order to be a function you should have one output for every input (or one y for every x). In this case there are multiple outputs for the number one which means the relationship isn't a function

4 0

3 years ago

Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the sur

Bad White [126]

Looks like you have most of the details already, but you're missing one crucial piece.

$\sigma$ is parameterized by

$\vec r(u,v)=\langle u\cos3v,u\sin3v,v\rangle$

for $0\le u\le7$ and $0\le v\le\frac{2\pi}3$ , and a normal vector to this surface is

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left\langle\sin3v,-\cos3v,3u\right\rangle$

with norm

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{\sin^23v+(-\cos3v)^2+(3u)^2}=\sqrt{9u^2+1}$

So the integral of $f(x,y,z)=x^2+y^2+z^2$ is

$\displaystyle\iint_\sigma f(x,y,z)\,\mathrm dA=\boxed{\int_0^{2\pi/3}\int_0^7(u^2+v^2)\sqrt{9u^2+1}\,\mathrm du\,\mathrm dv}$

6 0

3 years ago