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3 years ago

A marker was placed at an altitude that is 15 feet below sea level.

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

7 0

Answer:

-15 below sea level

-Hope this helps! :)

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Please help #12 10 points

eimsori [14]

1. add 40 to the 40 and 25
2. now you have
$\frac{5x {}^{2} }{5} = \frac{65}{5}$
3. then you have
${x}^{2} = 13$
4. then square root
${x}^{2} = 13$
5.then round answer to nearest tenth which is
$x = 3.61$

3 0

3 years ago

you pay 3.3 cents per minute for long-distance phone calls on your phone. your phone bill is $3.40 write and solve an equation t

MaRussiya [10]

If you would like to know the number of minutes of long-distance calls you made, you can calculate this using the following steps:

3.3 cents = $0.033
m = $3.40 / 3.3 cents
m = $3.40 / $0.033 = 3.40 / 0.033
m = 103.03 minutes

The correct result would be 103.03 minutes.

5 0

3 years ago

If u=(0,3), what is the magnitude of 3u

Rina8888 [55]

Answer:

9.

Step-by-step explanation:

3 * √)0^2 + 3^2)

= 3*3

= 9.

4 0

3 years ago

Find a function where f(0)=2 and f(1)=2

xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

$f(x) = x^2-x+2$

Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$