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The A triangle is congruent to G since they are symmetrical. (flipped over)

Triangle D is congruent to C since the 1/4 octagon is split in half, evenly.

40 responses saying "cat", 2 "4" bars,  so each bar is 40/4 /2= 10/2 = 5 people.

Dogs: 5 bars men, 6 women, so 11 * 5 = 55 dogs.

Hamster: 3men, 1 women, so 4*5 = 20 hammysters LOL(hamsters)

Other: 2 men 3 women so 5*5 = 25 othesrs.

If you don't understand anything feel free to ask!

How many speed limit signs are on this stretch of highway?

12 1/4 divided by 7/8 = (49/4)/(7/8) = 49/4 * 8/7 = 56/4 = 14

There's also a sign at the beginning, so 14+1 = 15 total signs.

If you don't understand anything feel free to ask.

I think you made a typo in your graph, you put  \(y-2x\le5\) instead of \(y-2x\le6\), but i fixed it and got 11.5 as answer.

Thanks for making the graph tho!(Idk how to use desmos)

The distance is 5/2.

The probability is 2/11.

Let x  be  the weight  of a sack of sugar

Let y  be  the weight  of  a sack of flour

We  have  these inequalities

2x + 3y ≤  40

y - 2x ≤ 6

See  the  graph here  :  https://www.desmos.com/calculator/rlbzlpxzgz

The    max  weight  of a sack of flour  occurs  at the "corner point" of  the  feasible region  =   11.25  lbs

Use   angle = arc length/radius   to get the answer in radians.  Multiply the result by 180/pi to get the result in degrees.

Let the sum of nine member (total) =x
and the age of old one=z
so its average 4 yrs before=(x+z)/10.
after 4 yrs let z be replaced by y.
so now avg=(x+4*10+y)/10

now (x+z)/10=(x+40+y)/10
so after solving it found
z=y+40.
so the older person is 40yrs older than young one

:) (sorry for the typo)

12 m/hr x  4.5 hr = 54 miles

x = 54 cos 143 =......          west  

y = 54 sin 143 = ......          north            assuming 0^o is North   ( I think the |41.3|  answer choices above  should be 43.1)

common ratio = 1/5     which is <1    so

Converges

geometric

arccos (7/18) = 67.12^o

the ladder is the hypotenuse and the side of the building and distance from the building are the legs , so the Pythagorean Theorem applies

9.8^2 = 3^2 + d^2       d = 9.33 m = distance up the side of the building

 if Jost is able to stand most of the way up the ladder    he will be able to reach a bit less than  9.33 + 1.2  + his arm length =~~ 10.5 m

   the top of the window is at    9.8 m  + 1.2 = 11 m  

    it is unlikely Josh will be able to clean to the top of the windows without a tool extension handle of some kind !

Go to the online graphing calculator, Desmos:  https://www.desmos.com/calculator 

Type in

to see the graph.

Which one? Or both. If you want you can answer one, you do not need to answer both. You don't have to answer at all if you do not want ! :)

The quadratic polynomial p(x) leaves a remainder of 1 on division by x - 1 or x - 2.  
The product of the roots of p(x) is 1.  
What is the sum of the roots of p(x)?

The quadratic polynomial is \( p(x)=ax^2+bx+c\)

\(\begin{array}{|lrcll|} \hline & p(x) &=& q_1(x)(x-1)+1 \quad | \quad p(x)=ax^2+bx+c \\ x=1: & ax^2+bx+c &=& q_1(x)(x-1)+1 \\ & a+b+c &=& q_1(x)\cdot 0 +1 \\ &\mathbf{ a+b+c} &=& \mathbf{1} \qquad (1) \\ \hline & p(x) &=& q_2(x)(x-2)+1 \quad | \quad p(x)=ax^2+bx+c \\ x=2: & p(x)=ax^2+bx+c &=& q_2(x)(x-2)+1 \\ & 4a+2b+c &=& q_2(x)\cdot 0+1 \\ & \mathbf{4a+2b+c} &=& 1 \qquad (2) \\ \hline \end{array}\)

We have roots \(r_1\) and \(r_2\)

\(\begin{array}{|rcll|} \hline p(x)=ax^2+bx+c &=& 0 \\ ax^2+bx+c &=& 0 \quad | \quad : a \\ x^2+\underbrace{\frac{b}{a}}_{=-(r_1+r_2)}x+\underbrace{\frac{c}{a}}_{=r_1r_2} &=& 0 \\ \hline r_1r_2= 1 &=& \frac{c}{a} \\ 1 &=& \frac{c}{a} \\ \mathbf{c} &=& \mathbf{a} \\ \hline -(r_1+r_2) &=& \frac{b}{a} \\ \mathbf{r_1+r_2} &=& \mathbf{-\frac{b}{a}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1): &\mathbf{ a+b+c} &=& \mathbf{1} \quad | \quad c= a \\ & a+b+a &=& 1 \\ & 2a+b &=& 1 \\ & \mathbf{b} &=& \mathbf{1-2a} \qquad (3) \\ \\ (2): & \mathbf{4a+2b+c} &=& \mathbf{1} \quad | \quad c= a \\ & 4a+2b+a &=& 1 \\ & 5a+2b &=& 1 \quad | \quad \mathbf{b=1-2a} \\ & 5a+2(1-2a) &=& 1 \\ & 5a+2-4a &=& 1 \\ & a+2 &=& 1 \\ & a &=& 1-2 \\ & \mathbf{ a } &=& \mathbf{-1} \\\\ (3): & \mathbf{b} &=& \mathbf{1-2a} \quad | \quad \mathbf{ a =-1} \\ & b &=& 1-2(-1) \\ & b &=& 1+2 \\ & \mathbf{ b } &=& \mathbf{3} \\\\ & \mathbf{c} &=& \mathbf{a} \quad | \quad \mathbf{ a =-1} \\ & \mathbf{c} &=& \mathbf{-1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{r_1+r_2} &=& \mathbf{-\frac{b}{a}}\quad | \quad b=3,\ a=-1 \\ r_1+r_2 &=& -\frac{3}{-1} \\ \mathbf{r_1+r_2} &=& \mathbf{3} \\ \hline \end{array}\)

The sum of the roots of \(p(x)= -x^2+3x-1\) is 3

Triangle ABC has vertices B and C on a semicircle centered at O, as shown, with AB tangent to the semicircle at B and AC intersecting the semicircle at point D.

If angle BAC = 78 degrees (in red) and angle ODC = 74  degrees (in green),

what is the measure of angle BCA (in blue) in degrees?

\(\begin{array}{|rcll|} \hline \mathbf{x+\varphi} &=& \mathbf{74^\circ} \quad | \quad \text{in triangle ODC} \\ \mathbf{\varphi} &=& \mathbf{74^\circ-x } \\\\ \mathbf{ 78^\circ +(90^\circ-\varphi) +x} &=& \mathbf{180^\circ} \quad | \quad \text{in triangle BAC} \\ 78^\circ +\Big(90^\circ-(74^\circ-x)\Big) +x &=& 180^\circ \\ 78^\circ +(90^\circ-74^\circ+x) +x &=& 180^\circ \\ 78^\circ + 90^\circ-74^\circ+x +x &=& 180^\circ \\ 94^\circ+2x &=& 180^\circ \quad | \quad :2 \\ 47^\circ+ x &=& 90^\circ \\ x &=& 90^\circ -47^\circ \\ \mathbf{x} &=& \mathbf{43^\circ} \\ \hline \end{array}\)

The measure of angle BCA is \(\mathbf{43^\circ}\)

If

\(\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{n} + \dfrac{1}{62} + \dfrac{1}{126} + \dfrac{1}{248} = \dfrac{495}{496}\)

 what is the value of \(n\)?

\(\begin{array}{|rcll|} \hline \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{n} + \frac{1}{62} + \frac{1}{126} + \frac{1}{248} &=& \frac{495}{496} \quad | \quad \times 2 \\ \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + \frac{2}{n} + \frac{2}{62} + \frac{2}{126} + \frac{2}{248} &=& \frac{495}{248} \\ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{2}{n} + \frac{1}{31} + \frac{1}{63} + \frac{2}{248} &=& \frac{495}{248} \\ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{2}{n} + \frac{1}{31} + \frac{1}{63} &=& \frac{495}{248} - \frac{2}{248} \\ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{2}{n} + \frac{1}{31} + \frac{1}{63} &=& \frac{493}{248} \quad | \quad \times 8 \\ 8 + \frac{8}{2} + \frac{8}{4} + \frac{8}{8} + \frac{16}{n} + \frac{8}{31} + \frac{8}{63} &=& \frac{493}{31} \\ 8 + 4 + 2 + 1 + \frac{16}{n} + \frac{8}{31} + \frac{8}{63} &=& \frac{493}{31} \\ 15 + \frac{16}{n} + \frac{8}{63} &=& \frac{493}{31} - \frac{8}{31} \\ 15 + \frac{16}{n} + \frac{8}{63} &=& \frac{485}{31} \\ \frac{16}{n} + \frac{8}{63} &=& \frac{485}{31} -15 \\ \frac{16}{n} + \frac{8}{63} &=& \frac{485-15*31}{31} \\ \frac{16}{n} + \frac{8}{63} &=& \frac{20}{31} \\ \frac{16}{n} &=& \frac{20}{31} - \frac{8}{63} \\ \frac{16}{n} &=& \frac{20*63-8*31}{31*63} \\ \frac{16}{n} &=& \frac{1012}{1953} \\ \frac{1}{n} &=& \frac{1012}{16*1953} \\ \frac{1}{n} &=& \frac{1012}{31248} \\ \frac{1}{n} &=& \frac{253}{7812} \\ \mathbf{n} &=& \mathbf{\frac{7812}{253}} \\ \hline \end{array}\)

See the answer here: https://web2.0calc.com/questions/the-table-shows-the-highest-daily-temperature-in#r1 to an almost identical question.  Only the temperature values in the table are different.  You should be able to use this to answer your question.

No problem!

I think I made a huge mistake. I've just realized that the dimension 4 is given at a certain angle. 

corrections:        V = 5 * 1.5 * sqrt(3)               V = [10 * 3 * 2sqrt(3)] / 8  

I apologize, Alan!!!  Your answer is, of course, correct!!! 

In triangle ABC, E is on AB such that AE:EB = 3:2, and D is on AC such that AD:DC = 1:4.  
F is the intersection of BD and CE.  
Find EF:FC.

\(\begin{array}{lclcll} \text{Let $\vec{AB}=\vec{b}$ } && \text{Let $\vec{AC}=\vec{c}$ }\\\\ \text{Let $\dfrac{AE}{AB}=\lambda=\dfrac{3}{5} $ } && \text{Let $1-\lambda=\dfrac{2}{5} $ }\\ \text{Let $\dfrac{AD}{AC}=\mu=\dfrac{1}{5} $ } && \text{Let $1-\mu=\dfrac{4}{5} $ } \\ \text{Let $\dfrac{EF}{EC}=x $ } && \text{Let $\dfrac{FC}{EC}=1-x $ }\\ \text{Let $\dfrac{EF}{FC}=\frac{x}{1-x} $ }\\\\ \text{Let $\dfrac{DF}{DB}=\eta $ }\\\\ \text{Let $\vec{AE}=\lambda\vec{b}$ } && \text{Let $\vec{AD}=\mu\vec{c}$ }\\\\ \text{Let $\vec{EC}=\vec{AC}-\vec{AE}=\vec{c}-\lambda\vec{b}$ } &&\text{Let $\vec{DB}=\vec{AB}-\vec{AD}=\vec{b}-\mu\vec{c}$ } \\\\ \text{Let $\vec{EF}=x\vec{EC}=x(\vec{c}-\lambda\vec{b})$ } &&\text{Let $\vec{DF}=\eta\vec{DB}=\eta(\vec{b}-\mu\vec{c})$ } \\ \end{array} \)

\(\begin{array}{|rcll|} \hline \vec{AE} + \vec{EF} &=& \vec{AD} + \vec{DF} \\ \lambda\vec{b} + x(\vec{c}-\lambda\vec{b}) &=& \mu\vec{c} + \eta(\vec{b}-\mu\vec{c}) \\ \lambda\vec{b} + x\vec{c}-x\lambda\vec{b} &=& \mu\vec{c} + \eta\vec{b}-\eta\mu\vec{c} \\ \vec{b}(\underbrace{\lambda-x\lambda- \eta}_{=0} ) &=& \vec{c} (\underbrace{\mu-\eta\mu- x}_{=0}) \\ \hline \lambda-x\lambda- \eta &=& 0 \\ \eta &=& \lambda-x\lambda \\ \mathbf{ \eta } &=& \mathbf{ \lambda(1-x) } \\ \hline \mu-\eta\mu- x &=& 0 \\ x &=& \mu-\eta\mu \\ x &=& \mu-\lambda(1-x)\mu \\ x &=& \mu-\lambda\mu(1-x) \\ x &=& \mu-\lambda\mu + \lambda\mu x \\ x - \lambda\mu x &=& \mu-\lambda\mu \\ x (1- \lambda\mu) &=& \mu(1-\lambda) \\ \mathbf{x} &=& \mathbf{\dfrac{ \mu(1-\lambda)}{1- \lambda\mu} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{ \mu(1-\lambda)}{1- \lambda\mu} } \\\\ x &=& \dfrac{ \dfrac{1}{5}* \dfrac{2}{5}}{1- \dfrac{3}{5}*\dfrac{1}{5} } \\\\ x &=& \dfrac{ \dfrac{2}{25}}{1- \dfrac{3}{25} } \\\\ x &=& \dfrac{ \dfrac{2}{25}}{ \dfrac{22}{25} } \\\\ x &=& \dfrac{2}{22} \\\\ \mathbf{x} &=& \mathbf{\dfrac{1}{11}} \\\\ \mathbf{1-x} = 1-\dfrac{1}{11} &=& \mathbf{\dfrac{10}{11}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{EF}{FC} &=& \dfrac{x}{1-x} \\\\ \dfrac{EF}{FC} &=&\dfrac{ \dfrac{1}{11} } { \dfrac{10}{11} } \\\\ \mathbf{\dfrac{EF}{FC}} &=& \mathbf{\dfrac{1}{10}} \\ \hline \end{array}\)

Imagine the angle was 89° (or even 90°). Do you think the volume would still be the same?

(Incidentally, I make mistakes all the time!  That's why it's incumbent on all posters to check the answers themselves.)

XY^2 = 392