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skelet666 [1.2K]
3 years ago
14

What other ways can you write 18:4

Mathematics
2 answers:
Crank3 years ago
7 0
18 to 4, and 18/4 those are the only two I could think of
Zepler [3.9K]3 years ago
5 0
18:4 can be written in fractional form = \frac{18}{4}
or as a mixed number;4 \frac{1}{2}
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brian borrowed $12,500 from the bank at an annual simple interest of 6.5%. if he repaid the loan in five years, how much did he
Lostsunrise [7]

Answer:

$16562.50

Step-by-step explanation:

3 0
2 years ago
1. The table shows a function.
Nataliya [291]

Answer: a) Linear function

 b) adding 8 each time x increases by 1

c) (0,13)


Step-by-step explanation:

For the given table, the rate of change is constant throughout the table.

The rate of change = \frac{change\ in\ y}{change\ in\ x}=\frac{21-13}{1-0}=8

Similarly we can check for every interval, the rate of change remains constant .

Thus, it is a linear function and the pattern we observe here is "adding 8 each time x increases by 1".

We know that the ordered pair of y intercept = (0,y)

In the table at x=0, y=13

hence, the y intercept of the given function is (0,13)

7 0
2 years ago
(9,-9)
erik [133]

Answer:

                               

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Solve x2 - 10x = -21
ipn [44]

Answer:

add 21 to both sides to get everything on one side

x^2 - 10x + 21 = 0

do factorization, guess and check

(x - 7) (x -3) = 0

x =7, and x =3

8 0
3 years ago
Read 2 more answers
G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =
Svetach [21]
Presumably you should be doing this using calculus methods, namely computing the surface integral along \mathbf r(u,v).

But since \mathbf r(u,v) describes a sphere, we can simply recall that the surface area of a sphere of radius a is 4\pi a^2.

In calculus terms, we would first find an expression for the surface element, which is given by

\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k
\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j
\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k
\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u

So the area of the surface is

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u
=-2\pi a^2(\cos\pi-\cos 0)
=-2\pi a^2(-1-1)
=4\pi a^2

as expected.
6 0
2 years ago
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