All Answers 358154 Answers

Why would you want to convert it to an improper fraction Chris?

I think this question could be better demonstrated on a number line.  

maybe I'll get up to organising  it one day!

the answer is just

$$-3\frac{1}{32}$$

I don't think so but thanks for the laughs today!  I only just saw your other prank!

Hi Chris,

Have you whited out your answer because there is 'There's nothing there...'? 

Good one!   lol

$$tan x\times cosx =\frac{sinx}{cosx}\times cos x = sinx\\\\

sinx=0.76\times 0.22$$

 $${\mathtt{0.76}}{\mathtt{\,\times\,}}{\mathtt{0.22}} = {\frac{{\mathtt{209}}}{{\mathtt{1\,250}}}} = {\mathtt{0.167\: \!2}}$$

 So $${\mathtt{sinx}} = {\mathtt{0.167\: \!2}}$$

This was my first thought but as Chris points out it is not correct because the question is nonsense.

If tanx = 0.76 then cosx CANNOT equal 0.22

Thanks Chris!

HA !!! .....that IS pretty funny......you and I were jousting at non-existent windmills (created by me, mostly)!!

I ALWAYS get in trouble when I start "assuming" things!!

I think we should wipe away our answers and pretend that it all never happened!!!

This is just another reason why huge and tiny numbers are so much better presented in scientific notation!

Thanks Chris but the first half of your first question is the most correct answer.

We were never told that it was velocity and time!

$$\frac{sin32}{15}=\frac{sin18.5}{a}\\\\
\frac{15}{sin32}=\frac{a}{sin18.5}\\\\
sin18.5\times\frac{15}{sin32}=a\\\\$$

$${\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{18.5}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{15}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{32}}^\circ\right)}}} = {\mathtt{8.981\: \!688\: \!659\: \!618\: \!659\: \!5}}$$

So $$a \approx 8.98\:\:\:\mbox{units}$$

Sorry...I typed in 14362 instead of 14632 !!  My bad.......

Note, that I said that "if" T ≥ 0......I realize that might not be the case.

Yeah....I think you're correct about the "negative" velocity thing...I hadn't considered that. I suppose we would have to know what the graph was intended to show......does the object stop and not move again, or does it stop and reverse course?? (Which would have to be the case if we were considering "negative" velocities.)

Since it's not stated, I'll have to assume the general case and that your answer is the better one........

$${\mathtt{2.38}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{1}}\right)}{\mathtt{\,\times\,}}{\mathtt{5}} = {\frac{{\mathtt{119}}}{{\mathtt{100}}}} = {\mathtt{1.19}}$$

or

$${\mathtt{2.38}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{5}}\right)} = {\mathtt{0.000\: \!023\: \!8}}$$

$${\mathtt{2.38}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}} = {\mathtt{238\,000}}$$

SO you have the minus sign in the wrong place, it should be out the front

$${\mathtt{\,-\,}}\left({\mathtt{2.38}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}}\right) = -{\mathtt{238\,000}}$$

$$5y-25-8y\\\\
=5y-8y-25\\\\
=-3y-25$$

Chris thinks you meant

$$(e^{-0.2}\times x) - (e^{-0.75}\times x)=0.25$$

Yes, this probably is what you meant.  Your title is missing a bracket so there is plenty of room for confusion!

I read in an extra x (without even realising) and Chris read in an extra bracket.  I now see your main text does have the bracket!

Is it possible to find the domain and range of V(T)=14632-(1504)T

I just thought i would think about and discuss Chris's answer a little more.

V=14632-1504T     If t is time then  $$t\ge 0$$     

Therefore the biggest V can be is 14632

I know it is 'normal' for only postive times to be considered but I don't know why you have rejected negative velocities.

also note that acceleration is -1504  (I think of this as a backward 'pull') Initially the velocity is positive but it is slowing down at approx t=9.73units of time the objects stops. [that is when 0=14632-1504T]. Where did you get 9.55 from Chris. Is my calculator broken?

$${\mathtt{0}} = {\mathtt{14\,632}}{\mathtt{\,-\,}}{\mathtt{1\,504}}{T} \Rightarrow \left\{ \begin{array}{l}\end{array} \right\}$$    Why didn't this work!

$${\frac{{\mathtt{14\,632}}}{{\mathtt{1\,504}}}} = {\frac{{\mathtt{1\,829}}}{{\mathtt{188}}}} = {\mathtt{9.728\: \!723\: \!404\: \!255\: \!319\: \!1}}$$

It then immediately begins moving backwards (negative velocity) It will then contiue in the negative direction getting faster and faster for ever.

So I think domain  $$[0,\inf)$$       Now I have fogotten the LaTex symbol for infinity!

and range $$(-\inf, 14632]$$

That is what I think anyway.  Plus, I have never completely understood why time is never considered in a negative direction.  Was there no yesterday???

Maybe you meant this:

e^(-.2) x  - e^(-.75) x  = .25 ??

I'm going to assume so......So we have

0.8187307530779818 x - 0.4723665527410147 x = .25

0.3463642003369671 x = .25        So x = .25 / 0.3463642003369671 = 0.7217836016446927047

I don't see any question here....

for this problem, the square root of 81 is 9 and -9 because: 

9 x 9 = 81 =) 

AND 

-9 x -9 = 81 =D (when you multiply two negative numbers you also get a positive) 

FINAL ANSWER: 

9 and -9 

$${\frac{{\mathtt{649.99}}}{{\mathtt{5}}}} = {\mathtt{129.998}}$$

that's $130 per month (but usually interest is also charged)

How about with the site calculator?

$${{\mathtt{e}}}^{\left(-{\mathtt{0.2}}\right)}{\mathtt{\,\times\,}}{\mathtt{0.25}}{\mathtt{\,-\,}}{{\mathtt{e}}}^{\left(-{\mathtt{0.75}}\right)}{\mathtt{\,\times\,}}{\mathtt{0.25}} = {\mathtt{0.086\: \!591\: \!050\: \!084\: \!241\: \!8}}$$

I am not sure if this is actually what you meant - you must be careful to check the brackets are in the right place.

(It is what you asked i believe)

In countries utilizing the "long scale," (Great Britain, most of continental Europe and Latin America) it has 18.

In countries utilizing the "short scale," (U.S. and Arabic-speaking countries) it has 12.

As they say on TV, consult your local listings........

This is just a linear function that has a slope of -1504 and a y intercept of 14362

It's domain and range are the same..... (-∞, ∞)  

In other words, all "Xs" and "Ys" are covered by this function

P.S. - This looks like a velocity function with a constant deceleration.

If T has to be ≥ 0,  then the domain is [0, ≈9.55] - since a "negative" velocity doesn't make sense, and the range would be [0, 14362] - In effect, the object stops after about 9.55 seconds, hours or whatever.

OK......  We have

Area =L x W = 72     and Perimeter = 2(L + W) = 36

Working with the Perimeter and dividing by 2 on each side...... we have

(L + W ) = 18    Now, subtract W from both sides......we have

L = 18 - W       And substituting for L in the area equation, we have

(18- W) x W = 72    or   18W - W² = 72    ... Add  W² to both sides and subtract 18W from both sides

This gives       0 = W² - 18W + 72

And factoring the right side we have 0 = (W-6) (W-12)

And setting each factor to 0 we get W = 6 or W = 12

So, depending on your orientation, either [W = 6ft and  L = (18 - 6) =12ft] or [W = 12ft and L = (18 - 12) = 6ft]

Either way, the garden is 6ft by 12ft

$${\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}{\mathtt{4}} = {\mathtt{\,-\,}}{\frac{{\mathtt{14}}}{{\mathtt{3}}}} = -{\mathtt{4.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$