radio

Multiply the three terms together to get the answer.

$${\mathtt{6}}{\mathtt{\,\times\,}}\left({\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right) = {\mathtt{70}}$$

70cm^3

Check this http://web2.0calc.com/questions/math-question-with-pic

First we find the size of the angle under the person on the platform. This is $${\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{79}} = {\mathtt{11}}$$, so 11 degrees.

Next we use trigonomerty to find the length of the hypotenuse the triangle with side 10m and angle 11 and 90.

10m = Adjacent

Angle = 11

Hypotenuse = x

Cos(11)=10/x

x=10/Cos(11)

$${\mathtt{x}} = {\frac{{\mathtt{10}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{11}}^\circ\right)}}} \Rightarrow {\mathtt{x}} = {\mathtt{10.187\: \!166\: \!949\: \!548\: \!654\: \!9}}$$

rounding to the nearest meter, x=10m so the answer is 10m

(i think).

So its 0.08=0.75x divided by 1 then take away 0.75?

Check this question. http://web2.0calc.com/questions/logarithm_5

Yes. 10^3 = 1000 * 2 = 2000

Im assuming you want to simplify.

6(5w-2x)-(4w-x)

Multiply first bracket by 6.

30w-12x-4w+x

26w-11x.

Note.-(4w-x) is turned into -4w+x when brackets are removed because it is the same as +(-1(4w-x)), so multiplying the -x by -1 turns it into +x or just x.

Find out how many times you need to divide by 10 to get the remaining digit to between 1 and <10.

Then put you answer in here:

(final digit)*10^(number of times needed to divide by 10 to get a digit on its own 1<= and <10)

8*10^11.

First, expand brackets, so

-7x-7-6=-41x+55

Next, combine like terms. To do this, add 41x to both sides of the equals sign, so you get 34x-7-6=55.

Then move the -7-6/ -13 over by adding 13 to both sides.

34x=55+13

34x=68

Divide both sides by 34 to get x on its own.

x=2

There you have it, the answer is 2

Alternatively, you could use trail and error. So:

$${{\mathtt{1}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{2}}}^{{\mathtt{3}}} = {\mathtt{9}}$$

Too low

$${{\mathtt{3}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{4}}}^{{\mathtt{3}}} = {\mathtt{91}}$$

Still too low

$${{\mathtt{6}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{7}}}^{{\mathtt{3}}} = {\mathtt{559}}$$

Too high

$${{\mathtt{5}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{6}}}^{{\mathtt{3}}} = {\mathtt{341}}$$

Bingo