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an integer is part of the real number system. natural number comes first: 1,2,3,4...... Integer is a number system that extends into the negatives and positives as well as 0

how you normally round, ex.:  12.346643242   take the number and round it to the nearest thousandths and you get 12.347 as ur answer

What i v = 3t - 4

$${\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{14}} = {\mathtt{23}}$$

Cool!whats the question btw?

Rose do you know whats the formula to find the area of the circle , if so can u tell me?

yes, it's the volume of water first, then hydrogen. hardly anyone gets that right.

well, basically, it's the volume divided by the mass, then times by the volume of hydrogen.

1 pound . Because 12 eggs weigh 1.5 pounds and 8 is two thirds of 12 , so you take away one third of the weigh which is half a pound :)

yay caroline:) BESTIES:)

use mathway its an app i used to cheat my way thrugh school

work out what?

its window.

or commonly known as

the number 2.

Melosy i'll tell you that some time esle! a long story!

melody your an expert in LaTex i wish i could even be!

You are very welcome :)

Thanks so much Melody!

good bye!

I teleported the butter

do you mean

(48x^8-14x^6)/(7x)

The ^ means power of

which is 

$$\\\frac{48x^8-14x^6}{7x}
=\frac{48x^7-14x^5}{7}\\\\$$

θ is the angle whose cosine is 3*√(6)/14 so

θ = cos^-1(3*√(6)/14)

$${\mathtt{theta}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}}}}{{\mathtt{14}}}}\right)} \Rightarrow {\mathtt{theta}} = {\mathtt{58.339\: \!117\: \!225\: \!405^{\circ}}}$$

180° is pi radians, so in radians we have

$${\mathtt{theta}} = {\frac{{\mathtt{58.339\: \!117\: \!225\: \!405}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}} \Rightarrow {\mathtt{theta}} = {\mathtt{1.018\: \!209\: \!678\: \!290\: \!256\: \!2}}$$

or θ ≈ 1.02 radians

This seems to refer to the well-known right-angled triangle with sides BC = 3, AC = 4, AB = 5, in which case cos(CAB) = 4/5 (=0.8)

1. Find the distance from A to the spot on the ground immediately below the balloon (i.e. point (0, 0, 0)) - I'll call this spot E.  Use Pythagoras theorem to do this.  AE = √(5² + 9²).

2.  The force along EA is F_EA = F_ADcos(A), where A = tan^-1(20/AE)

3.  Resolve F_EA along the x and y directions.  F_EAx = F_EA*5/AE,  F_EAy = -F_EA*9/AE

4.  Replace F_EA in step 3 by F_ADcos(A).

5.  Do the same for the forces F_BD and F_CD.

6.  Resolve F_AD, F_BD, and F_CD along the z-direction.

7.  Balance the forces in the x-direction.  Balance the forces in the y-direction.  Balance the forces in the z-direction (don't forget the vertical force on the balloon).  This should give you three equations for the three unknown forces F_AD, F_BD, and F_CD.

The height of a certain object is given by h(t)= t² + 1 . Find the average velocity over the time period [4, 5].

Hint: find the average rate of change (AROC), or rise/run

When t=4         height =17

When t=5          height = 26

Average rate of change = (26-17)/(5-4)  

Let F_AB be the force in cable AB and F_BC be the force in cable BC.  Let θ_A be the angle between cable AB and the horizontal and θ_C be the angle between cable BC and the horizontal.

Horizontal force balance

F_ABcos(θ_A) = F_BCcos(θ_C)          ...(1)

Vertical force balance

F_ABsin(θ_A) = mg + F_BCsin(θ_C)  ...(2)

m = 2000kg; g = 9.81m/s².

We also have that θ_A = tan^-1(5.6/2.2) and θ_C = tan^-1(2.4/5.6)

Replace F_BC in (2) by using (1)

F_ABsin(θ_A) = mg + F_ABcos(θ_A)sin(θ_C)/cos(θ_C)

Rearrange to get F_AB

F_AB = mg/(sin(θ_A) - cos(θ_A)sin(θ_C)/cos(θ_C))

$${\mathtt{FAB}} = {\frac{{\mathtt{2\,000}}{\mathtt{\,\times\,}}{\mathtt{9.81}}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.6}}}{{\mathtt{2.2}}}}\right)}\right)}{\mathtt{\,-\,}}{\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.6}}}{{\mathtt{2.2}}}}\right)}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2.4}}}{{\mathtt{5.6}}}}\right)}\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2.4}}}{{\mathtt{5.6}}}}\right)}\right)}}}\right)}} \Rightarrow {\mathtt{FAB}} = {\mathtt{25\,347.418\: \!086\: \!936\: \!201\: \!807\: \!8}}$$

FAB ≈ 25.347 kN

Substitute this back into (1) to get F_BC

F_BC = F_ABcos(θ_A)/cos(θ_C)

$${\mathtt{FBC}} = {\frac{{\mathtt{25\,347.418}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.6}}}{{\mathtt{2.2}}}}\right)}\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2.4}}}{{\mathtt{5.6}}}}\right)}\right)}}} \Rightarrow {\mathtt{FBC}} = {\mathtt{10\,083.657\: \!337\: \!858\: \!801\: \!218\: \!7}}$$

FBC ≈ 10.083 kN

$$\begin{tabular}{cccccc}
&&\;x^2&-11&+28&
&&&||&||&||&||&
x&-2&\|\;x^3&-13x^2&+50x&-56\\
&&\;x^3&-2x^2&&&
&&||&||&||&||&
&&&-11x^2&+50x&-56&\\
&&&-11x^2&+22x&&
&&&||&||&||&
&&&&+28x&-56&\\
&&&&+28x&-56&
&&&&||&||&
&&&&&0&
\end{tabular}$$

\begin{tabular}{cccccc} &&\;x^2&-11&+28& &&&||&||&||&||& x&-2&\|\;x^3&-13x^2&+50x&-56\\ &&\;x^3&-2x^2&&& &&||&||&||&||& &&&-11x^2&+50x&-56&\\ &&&-11x^2&+22x&& &&&||&||&||& &&&&+28x&-56&\\ &&&&+28x&-56& &&&&||&||& &&&&&0& \end{tabular}