The size of the regions does not matter here. What matters is the angle measurement, or, equivalently, what part of a circle each sector is.

Two of the blue sectors are each one third of one quarter-circle, and thus are

$\displaystyle \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$

of one circle. 

The other two blue sectors are each one fourth of one quarter-circle, and thus are

$\displaystyle \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

of one circle. 

Therefore, the total angle measure comprises

$\displaystyle \frac{1}{16} + \frac{1}{16} + \frac{1}{12} + \frac{1}{12}$

$\displaystyle =\frac{3}{48} + \frac{3}{48} + \frac{4}{48} + \frac{4}{48}$

$\displaystyle = \frac{14}{48} = \frac{7}{24}$

of a circle. This makes $\displaystyle \frac{7}{24}$ the correct probability. As odds, this translates to 

$\displaystyle (24 -7): 7$, or $\displaystyle 17:7$ odds against the spinner landing in blue.

The size of the regions does not matter here. What matters is the angle measurement, or, equivalently, what part of a circle each sector is.

The purple region is one third of one quarter of a circle, or, equvalently,

$\displaystyle \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$ 

of a circle, so its central angle is $\displaystyle \frac{1}{12}$ of the total measures of the angles of the sectors. This makes $\displaystyle \frac{1}{12}$ the probability of the spinner stopping inside the purple region; this translates to

$\displaystyle (12-1):1$ or $\displaystyle 11:1$ odds against this occurrence.

$\displaystyle \angle A$ and $\displaystyle \angle B$ are a pair of adjacent angles of the parallelogram, and as such, they are supplementary - that is, their degree measures total 180. Therefore, 

$\displaystyle m \angle A + m \angle B = 180 ^{\circ }$

$\displaystyle m \angle B = 180 ^{\circ } - m \angle A$

$\displaystyle = 180 ^{\circ } - (t+ 56 )^{\circ }$

$\displaystyle = (180 - 56 - t )^{\circ }$

$\displaystyle = (124-t )^{\circ }$

The sum of the measures of the angles of a triangle is 180 degrees, so solve for $\displaystyle t$ in the equation:

$\displaystyle m \angle A + m \angle B + m \angle C = 180^{\circ }$

$\displaystyle (3t-14) + (2t+16) + 5t = 180$

$\displaystyle 10t+ 2 = 180$

$\displaystyle 10t+ 2- 2 = 180 - 2$

$\displaystyle 10t = 178$

$\displaystyle 10t \div 10 = 178 \div 10$

$\displaystyle t = 17.8$

$\displaystyle m \angle A = (3t-14)^{\circ } = (3 \cdot 17.8-14)^{\circ } = (53.4-14)^{\circ } = 39.4^{\circ }$

$\displaystyle m \angle B = (2t+16)^{\circ } = (2 \cdot 17.8 + 16)^{\circ }= (35.6+ 16) ^{\circ }= 51.6^{\circ }$

$\displaystyle m \angle C =( 5t)^{\circ } = ( 5 \cdot 17.8)^{\circ } = 89 ^{\circ }$

All three angles measure less than 90 degrees and are therefore acute angles; that makes $\displaystyle \bigtriangleup ABC$ an acute triangle.

The total degrees of the angles in a triangle are $\displaystyle 180$.  

Since it is a right triangle, one of the three angles must be $\displaystyle 90.$ 

That leaves you with $\displaystyle 90$ for the other two angles $\displaystyle (180-90=90)$.  

If they are equal, you just divide the remaining degrees by $\displaystyle 2$ to get $\displaystyle 90/2=45$.

A right triangle has one 90 degree angle and all three angles must equal 180 degrees.

To find the answer, just subtract the two angles you have from the total to get 

$\displaystyle 180^\circ-\textup{Angle 1}-\textup{Angle 2}=\textup{Angle 3}$

The angle we have are,

$\displaystyle \\ \textup{Angle 1}=90^\circ \\ \textup{Angle 2}=45^\circ$.

Substituting these into the formula results in the solution.

$\displaystyle 180^\circ-90^\circ-45^\circ=45^\circ$.

To find the speed of the deer, you must have the distance traveled and the time.

The distance is found using the Pythagorean Theorem:

$\displaystyle A^2 + B^2 = C^2$

$\displaystyle C = {\sqrt{A^2+ B^2}}$

$\displaystyle C = \sqrt{30^2 + 40^2}$

$\displaystyle C = \sqrt{2500} = 50$

The answer must be in miles per hour so the total miles are divided by the hours to get the final answer:

$\displaystyle \frac{miles}{hour} = \frac{50}{8} = 6.25 mph$

The coordinate length between you and your school is equivalent to the hypotenuse of a right triangle with sides of 3 and 4 units:

$\displaystyle x^2=3^2+4^2$

$\displaystyle x^2=9+16$

$\displaystyle x^2=25$

$\displaystyle x=5$

The distance is 5 coordinate lengths, and each coordinate length corresponds to 1.3 miles of distance, so

$\displaystyle 5\times 1.3 =6.5\hspace{1 mm}miles$

The coordinates of a point are determined by the distance from the origin. The first point in the ordered pair is the number of units to the left or right of the origin. Negative numbers indicate the number of units to the left while positive numbers indicate the number of units to the right. The second number indicates the number of units above or below the origin. Positive numbers indicate the number of units above while negative numbrs indicate the number of units below the origin. The vertices of the square are:
$\displaystyle \small (-1,2); (-1,4); (1,2); (-1,4)$

A point is located on the $\displaystyle y$-axis if and only if it has $\displaystyle x$-coordinate (first coordinate) 0. Of the five choices, only $\displaystyle (0,-30)$ fits that description.