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I'm not the best at this, but this is my attempt:

150  =  2 x 3 x 5 x 5  --  so every multiply must have at least 1 two, 1 three, and 2 fives.

So, from the 2¹⁰ (which consists of 10 twos),

you can select either 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 twos  -- 10 different selections.

From 3¹⁴ (which consists of 14 threes), 

you can select either 1, 2, 3, ... 14 threes  --  14 different selections.

From 5⁸, you must select at least 2 fives, but it could also be 3, 4, 5, 6, 7, or 8 fives

  --  7 different selections.

Multiplying 10 x 14 x 7, I get 980 different ways.

Какое уравнение

1728

Use the cosine rule twice.

First in the triangle ABC to calculate the angle at A. (I got cos(A) = 1/10).

Second in the triangle ADE to calculate DE. 

I got DE = 18, but that should be checked.

Let a be the weight of Large box

and Let b be the weight of Small box

therefore, the equation will be

   =>  5a + 6b = 175kg ----------(1)

         3a + 2b = 83kg ------(2)

  now, Multiplying  equation (2) by 3 on both sides

we get,

       3a*3  + 2b*3 = 83*3

   =>  9a + 6b = 249 -----(3)

now subtracting equ. 2 and equ. 3

we get,

        

        5a + 6b = 175

  (-) 9a(-)+6b =(-)249

-------------------------   

    =>  -4a = -74

                 a =  -74/-4 

 

               --------------

              /   a=  18.5   /

                -------------

Therefore the weight of the large box is 18.5 kg

now ,

subtituting value of a in equation b

we get,

       3a + 2b = 83

=> 3*(18.5) + 2b = 83

        55.5 + 2b = 83

          2b = 83 - 55.5

          2b = 27.5

            b = 27.5 / 2

        -----------------

       /   b = 13.75    /

        -----------------

              Therefore the weight of the small box is 13.75kg

and the weight of large box is 18.5kg 

Points \(A\) and \(B\) are on side \(\overline{YZ}\) of rectangle \(WXYZ\) such that \(\overline{WA}\) and \(\overline{WB}\) trisect \(\angle ZWX\).
If \(BY = 3\) and \(AZ = 6\), then what is the area of rectangle \(WXYZ\)?

\(\begin{array}{|lrcll|} \hline (1) & \tan(A) &=& \dfrac{6}{y} \\\\ (2) & \tan(2A) &=& \dfrac{y}{6} \\\\ \hline (1)\times (2): & \tan(A)\times \tan(2A) &=& \dfrac{6}{y}\times \dfrac{y}{6} \\\\ & \tan(A)\times \tan(2A) &=& 1 \quad | \quad \tan(2A)=\dfrac{2\tan(A)} {1-\tan^2(A)} \\ & \tan(A)\times\dfrac{2\tan(A)} {1-\tan^2(A)} &=& 1 \\\\ & \dfrac{2\tan^2(A)} {1-\tan^2(A)} &=& 1 \\\\ & 2\tan^2(A) &=& 1-\tan^2(A) \\ & 3\tan^2(A) &=& 1 \\\\ & \tan^2(A) &=& \dfrac{1}{3} \\\\ & \tan(A) &=& \mathbf{ \dfrac{1}{\sqrt{3}} } \\ \hline (1) & \tan(A) &=& \dfrac{6}{y} \\\\ & \dfrac{1}{\sqrt{3}} &=& \dfrac{6}{y} \\\\ & \sqrt{3} &=& \dfrac{y}{6} \\\\ & \mathbf{y} &=& \mathbf{6\sqrt{3}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (3) & \tan(A) &=& \mathbf{\dfrac{y}{x-3}} \\\\ & x-3 &=& \dfrac{y}{\tan(A)} \\\\ & x &=& 3+\dfrac{y}{\tan(A)} \\\\ & x &=& 3+\dfrac{6\sqrt{3}}{\dfrac{1}{\sqrt{3}}} \\\\ & x &=& 3+ 6\sqrt{3}\sqrt{3} \\\\ & x &=& 3+ 18 \\\\ & \mathbf{x} &=& \mathbf{21} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{The area of rectangle }~ WXYZ &=& xy \\ &=& 21*6\sqrt{3} \\ &=& \mathbf{126\sqrt{3}} \\ &=& \mathbf{218.238401754} \\ \hline \end{array}\)

Points \(A\), \(B\), and \(C\) are on sides\( \overline{WX}\), \(\overline{YZ}\), and \(\overline{XY}\) of rectangle \(WXYZ\) as shown below such that
\(XA = 4\), \(YB = 18\), \(\angle ACB= 90^\circ\), and \(YC = 2XC\).
Find \(AB\).

\(\begin{array}{|rcll|} \hline \tan(A) = \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{XC}{9} \\\\ 36 &=& XC^2 \\\\ 6 &=& XC \\ \mathbf{XC} &=& \mathbf{6} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x^2 &=& (3XC)^2 + 14^2 \\ x^2 &=& (3*6)^2 + 14^2 \\ x^2 &=& 18^2 + 14^2 \\ x^2 &=& 324 + 196 \\ x^2 &=& 520 \\ x^2 &=& 4*130 \\ x &=& 2\sqrt{130} \\ \mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array}\)

\(\mathbf{AB \approx 22.8}\)

deledet

What was wrong with geno3141's answer ???

He finished his answer with this in the last line (under your first post)

  -17t + 30  =  -18t + 40

And said to you " And continue solving". You could't finish the above line by putting "t" on one side ??

-17t + 18 t =40 - 30

t = 5, which is your answer !!!!!!!!!!!!!!. 

T = Tanker(empty)
F =Fuel
[T + 0.25F]=2500.........(1)
[T + 0.75F]=3700.........(2), solve for F, T
F =2400 tons - weight of fuel when Tanker is full.
T =1900 tons - weight of Tanker when empty.

As follows:

Find the area of a triangle with side lengths 4/3, 2 and 8/3.

Hello Gast!

\(s=\frac{a+b+c}{2}\)

\(s=\frac{4}{3\cdot 2}+\frac{2}{2}+\frac{8}{3\cdot 2}=\frac{18}{6}\\ s=3\)

Heron

\(A=\sqrt{s(s-a)(s-b)(s-b)}\\ A=\sqrt{3(3-4/3)(3-2)(3-8/3)}\\ A=\sqrt{3\cdot 5/3\cdot 1\cdot 1/3}\\ A=\sqrt{\frac{15}{9}}\)

\(A=1.291\)

  !

Oh! That does work ty!!

On Tuesday, I worked t+1 hours and earned 3t-3 dollars per hour.
My friend Andrew worked 3t-5 hours but only earned t+2 dollars an hour.
At the end of the day, I had earned two dollars more than he had. What is the value of t?

\(\begin{array}{|rcll|} \hline (3t-3)\dfrac{\text{dollars}}{\text{hours}}\times (t+1)\text{hours} &=& $2+ (t+2)\dfrac{\text{dollars}}{\text{hours}}\times (3t-5)\text{hours} \\\\ (3t-3)(t+1) \text{dollars} &=& $2+ (t+2)(3t-5) \text{dollars} \\\\ (3t-3)(t+1)&=& 2+ (t+2)(3t-5)\\ 3(t-1)(t+1)&=& 2+ (t+2)(3t-5)\\ 3(t^2-1)&=& 2+ (t+2)(3t-5)\\ 3t^2-3 &=& 2+ (t+2)(3t-5)\\ 3t^2-3&=& 2+ 3t^2-5t+6t-10 \\ -3&=& 2+ -5t+6t-10 \\ -3&=& 2+t-10 \\ -3&=& t-8 \\ t &=& 8-3 \\ \mathbf{t} &=& \mathbf {5} \\ \hline \end{array}\)

DELETED

Try writing an eqaution for t, or see if guess and check will work

In an election, 1200 total votes were cast for the 3 candidates.
The second-place candidate received 125 fewer votes than the winner and 200 votes more than the third-place candidate.
How many votes did the winner receive?

\(\text{Let winner candidate = c_1 } \\ \text{Let second-place candidate = c_2 } \\ \text{Let third-place candidate = c_3 } \)

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{c_2} &=& \mathbf{c_1 - 125} \\ \hline (2) & c_2 &=& c_3 + 200 \\ & c_3 &=& c_2 - 200 \quad | \quad \mathbf{c_2=c_1 - 125} \\ & c_3 &=& c_1 - 125 - 200 \\ & \mathbf{c_3} &=& \mathbf{c_1 - 325} \\ \hline (3) &c_1+ c_2+c_3 &=& 1200 \quad | \quad \mathbf{c_2=c_1 - 125},\ \mathbf{c_3=c_1 - 325} \\ &c_1+ (c_1 - 125) + (c_1 - 325) &=& 1200 \\ &3c_1- 125 - 325 &=& 1200 \\ &3c_1- 450 &=& 1200 \\ &3c_1 &=& 1200 + 450 \\ &3c_1 &=& 1650 \quad | \quad : 3 \\ & \mathbf{c_1} &=& \mathbf{550} \\ \hline \end{array}\)

The winner did receive 550 votes?

thank u so much

In the prime factorization of \(109!\), what is the exponent of \(3\)?

(Reminder: The number \(n!\) is the product of the integers from \(1\) to \(n\).

For example, \(5!=5\cdot 4\cdot3\cdot2\cdot 1= 120\).)

\(\begin{array}{|rcll|} \hline \dfrac{109}{3^1} &=& \lfloor 36.3333333333 \rfloor \\ &=& 36 \\ \hline \dfrac{109}{3^2} &=& \lfloor 12.1111111111 \rfloor \\ &=& 12 \\ \hline \dfrac{109}{3^3} &=& \lfloor 4.03703703704 \rfloor \\ &=& 4 \\ \hline \dfrac{109}{3^4} &=& \lfloor 1.34567901235 \rfloor \\ &=& 1 \\ \hline \dfrac{109}{3^5} &=& \lfloor 0.44855967078 \rfloor \\ &=& 0 \\ \ldots \\ &=& 0 \\ \hline \text{sum} && 36+12+4+1= \mathbf{53} \\ \hline \end{array} \)

\(\mathbf{109!} = 2^{104}×\mathbf{3^{\color{red}53}}×5^{25}×7^{17}×11^9×13^8×17^6×19^5×23^4×\cdots\)

Thank you so much for explaining this to me!

Let \(a\) and \(b\) be positive real numbers such that \(a^b = b^a,\ b = 9a\),
Then \(a\) can be expressed in the form \(sqrt(n,m)\) where \(m\) and \(n\) are positive integers
and \(n\) is as small as possible.

Find \(m + n\)

\(\begin{array}{|rcll|} \hline \mathbf{a^b} &=& \mathbf{b^a} \quad | \quad \mathbf{b = 9a} \\\\ a^{9a} &=& (9a)^a \\\\ a^{9a*\frac{1}{a}} &=& \left(9a\right)^{a*\frac{1}{a}} \\\\ a^{9} &=& 9a \quad | \quad *(a^{-1}) \\\\ a^{9}a^{-1} &=& 9a^1a^{-1} \\\\ a^{9-1} &=& 9a^{1-1} \\\\ a^{8} &=& 9a^0 \quad | \quad a^0 = 1 \\\\ a^{8} &=& 9 \\\\ a^{8*\frac{1}{8}} &=& 9^{\frac{1}{8}} \\\\ a &=& 9^{\frac{1}{8}} \\\\ a &=& (3^2)^{\frac{1}{8}} \\\\ a &=& 3^{2*\frac{1}{8}} \\\\ a &=& 3^{\frac{2}{8}} \\\\ a &=& 3^{\frac{1}{4}} \\\\ \mathbf{a} &=& \mathbf{\sqrt[4]{3}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline m+n &=& 3+4 \\ \mathbf{m+n} &=& \mathbf{7} \\ \hline \end{array}\)

Post New Answer

\(\log_{10}(x) = 3 + \log_{10}(y)\)

find \(\dfrac{x}{y}\)

Formula:

\(\boxed{\log\left({\dfrac{x}{y}}\right) = \log(x)-\log(y)}\)

\(\begin{array}{|rcll|} \hline \log_{10}(x) &=& 3 + \log_{10}(y) \quad | \quad -\log_{10}(y) \\\\ \log_{10}(x) -\log_{10}(y) &=& 3 \\\\ \log_{10}\left({\dfrac{x}{y}}\right) &=& 3 \\\\ \dfrac{x}{y} &=& 10^3 \\\\ \mathbf{ \dfrac{x}{y} } &=& \mathbf{1000} \\ \hline \end{array}\)