proyaop

First we make the inequality clearer, this is what it looks like:

${t\over3} - 5 < t - 2$

Step 1 is to make it neater, so we multiply both sides of the inequality by 3.

$t - 15 < 3t - 6$

Then we isolate the variable by adding 6 to both sides of the inequality, then subtracting $t$ from both sides of the inequality.

$t - 9 <3t$

$-9<2t$

The last step is to divide both sides by 2 to get the interval for $t$.

-4.5 < t

4.5 can be written as 9/2, but there you have it, the interval for t!

To get the equidistant point (midpoint) of a line segment on a graph, all you need to do is to take the absolute value of the difference in x1 and x2, and y1 and y2, and divide it by 2, then that is the coordinate of your point.

Since the problem is only asking for the x coordinate, we can take the x coordinate from point A and divide it by 2, getting the x coordinate of the midpoint is 1.

That is the more detailed explanation for why the x coordinate is 1.

First we list the two equations.

$x + 3y = -4$                    (1)

$3x - 7y = 36 + 5x - 2y$ we must simplify the second equation to get:

$-2x - 5y = 36$               (2)

Now we have a system of equations that looks much nicer. On to the math:

First we multiply equation(1) by -2 to get:

$-2x - 6y = 8$                  (3) We will call this equation 3 for the sake of clarity.

Next we subtract equation(3) from equation(2) to cancel out variable x.

$-2x - 5y - (-2x - 6y) = 36 - 8$ This simplifies to:

$y = 28$

Now we can plug the value of $y$ back into equation(1) to get $x$:

$x + 84 = -4$ Then we subtract 84 from both sides of the equation to get x:

$x = -88$

Now we have the ordered pair of (x,y)

$x = -88$

$y = 28$

Because ABCD is a rhombus, then AB = BC = CD = AD.

AB = BC = CD = AD = 37 units.

One of rhombus ABCD's diagonals has length 28. We can divide 28 by 2 to get 14, which is the leg of a right triangle when rhombus ABCD is split into 4 congruent triangles.

The hypotenuse is 37, and the leg is 14, so that means the other leg is:$\sqrt{37^2 - 14^2}$  = $\sqrt{1173}$. Ew that looks ugly, but it is correct. That means the other diagonal has length $2\sqrt{1173}$. 

Since the area of a rhombus is one diagonal length times the other diagonal length over 2, then the area of this rhombus would be $28\sqrt{1173}$ units squared.

This figure is not drawn to scale. 

We can exclude BD and AB because using the Hinge Theorem, AD is the smallest line segment out of triangle ADB.

Angle C = 65, Angle CDA = 60, and Angle DAC = 55.

Then using the Hinge Theorem, CD is the included side of the two largest angles in triangle ACD:

Therefore CD is the shortest side.

We can list them since we know that a multiple of 3 has to have a digit sum that is divisible by 3:

24

42

222

444

That's it... So we have 4 positive multiples of 3 less than 1000 using only the digits 2 and/or 4.

pr

We can first label the radii of each of the circles to a variable.

Radius of circle A = x

Radius of circle B = y

Radius of circle C = z

Now we can write some equations:

$x + y = 14$

$y + z = 18$

$x + z = 28$

Then we add up the three equations and divide by two to get:

$x + y + z = 30$

We take the equation above us (equation4) and subtract it from equation2 to get x = 12.

We take the equation above us (equation4) and subtract it from equation3 to get y = 2.

We take the equation above us (equation4) and subtract it from equation1 to get z = 16.

So the radius of circle A is 12

So the radius of circle B is 2

So the radius of circle C is 16

GEMETRY

Given the angles, we can conclude that triangles TSR ~ triangle TRQ ~ triangle RQP. 

Using base angles theorem, these 3 triangles are all isosceles, and QR = 8. That means QT = 8, and QS = 2 so ST = 6.

The ratio of QS / ST is also the ratio of PT / TR. Which is 1 : 3

Now after we have concluded the information above, we can set segment TR as $x$. 

Since we know there are similiar triangles, and we know TS = 6 and QR = 8, then we can say ${x\over6} = {8\over x}$. 

Solving for x we get $x = 4\sqrt{3}$. PT = one third of TR, so PT = $4\sqrt{3}\over3$.

PQ = PT + PR.

Plugging in the values we get:

PQ = $16\sqrt{3}\over3$

gemetry

To graph it out, we would see a right triangle with legs of length $3\sqrt{5}$ and $d + 6$. If the hypotenuse is $3d$, then we can use the pythagorean theorem to get this equation:

$(3\sqrt{5})^2 + (d + 6)^2 = (3d)^2$

Simplified:

$45 + d^2 + 36 + 12d = 9d^2$

Combine like terms and subtraction:

$-8d^2 + 12d + 81 = 0$

Now we have a quadratic, we can apply the quadratic formula $d$ = $-b {+\over} \sqrt{b^2 - 4ac}\over2a$ where a is the coefficient of $d^2$, b is the coefficient of $d$, and c is the constant of the equation.

Plugging in the values, we get:

$-3 {+\over} 3\sqrt{19}\over-4$ = d

Since we need the smallest value of d, and d can't be a negative distance away from something, then we will use the subtraction operation. This simplifies to:

$d = {3 + 3\sqrt{19}\over4}$