hectictar

Since the range is 10, the first and last scores must be different. And we only have 4 scores total. So in order for there to be a mode, and for that mode to be 44, it must be that the middle two scores are both 44 (just like you have already put )

Then let's let  a  be the first (lowest) score, and let  d  be the last (highest) score.

Since the mean is 45, we know:

(a + 44 + 44 + d) / 4   =   45

(a + 88 + d) / 4   =   45

                                       Multiply both sides of the equation by 4

a + 88 + d   =   180

                                       Subtract  88  from both sides of the equation.

a  +  d   =   92

                                       Subtract  d  from both sides of the equation.

a   =   92  -  d

Since the range is 10, we know:

d  -  a   =   10

                              Substitute  92 - d  in for  a

d - (92 - d)  =  10

                              Distribute  -1  to both terms in parenthesees

d - 92 + d  =  10

                              Combine like terms

2d  - 92  =  10

                              Add  92  to both sides

2d  =  102

                              Divide both sides by  2

d  =  51

Now we can find  a  using this value of  d:

a   =   92  -  d

                              Substitute  51  in for  d

a   =   92  -  51

a   =   41

Maybe Nerdo meant that his own statement was off topic.

The two larger semicircles make one full circle with a diameter of 16 cm.

The two smaller semicircles make one full circle with a diameter of 8 cm.

So altogether we have a rectangle, a larger circle, and a smaller circle.

The total area is the sum of the areas of these three shapes.

So let's work on finding the area of each shape.

(Note that all areas will be in square cm)

area of larger circle   =   π ( radius )²   =   π ( 16 / 2  )²   =   π (  8  )²   =   64π

area of smaller circle   =   π ( radius )²   =   ...   (Can you finish this part? )

area of rectangle   =   length * height   =   16 * 8   =   128

Then add the three areas together to find the total area.

total area   =   area of rectangle  +  area of larger circle  +  area of smaller circle

total area   =   128   +   64π   +   ...   =   ?

We are given that the solids are made from the same material, which means they have the same density.

Since   mass   =   density * volume

mass_A / mass_B   =   (density_A * volume_A) / (density_B * volume_B)

                                                                                                           density_A = density_B   so we can cancel them out

mass_A / mass_B   =   volume_A / volume_B

So here we have shown that the ratio of the masses is equal to the ratio of their volumes.

Now we need to be careful and remember that the ratio of their volumes is not equal to the ratio of their surface areas.

The ratio of their volumes is equal to the ratio of their surface areas raised to the power of 3/2

1.

Remember:       distance   =   rate * time

Since he was going at two different rates for each part of the trip, lets first calculate the distance he flew during the first part of the trip and then we can calculate the distance he flew during the second part of the trip. The total distance traveled will be the sum of these two distances.

distance₁   =   rate₁ * time₁   =   680 mi/hr  *  (3.8 / 2) hr   =   680 mi/hr   *   1.9 hr   =   1292 mi

distance₂   =   rate₂ * time₂   =   740 mi/hr  *  (3.8 / 2) hr   =   740 mi/hr   *   1.9 hr   =   1406 mi

total distance   =   distance₁  +  distance₂   =   1292 mi  +  1406 mi   =   ?

2.

By solving the first formula for time,  we get:       time   =   distance / rate

Again let's split this into two parts for each part of the trip.

For the 1st part..... the distance is one third of the total distance, and the rate is 1032 km/hr.

For the 2nd part.... the distance is two thirds of the total distance, and the rate is 696 km/hr.

\(\text{time}_1\ =\ \dfrac{\text{distance}_1}{\text{rate}_1}\ =\ \dfrac{\frac13\cdot1860\ \text{km}}{1032\ \frac{\text{km}}{\text{hr}} } \ =\ \dfrac{620\ \text{km}}{1032\ \frac{\text{km}}{\text{hr}} }\ =\ \dfrac{620}{1032}\ \text{hr}\)

\(\text{time}_2\ =\ \dfrac{\text{distance}_2}{\text{rate}_2}\ =\ \dfrac{\frac23\cdot1860\ \text{km}}{696\ \frac{\text{km}}{\text{hr}} }\ =\ \dfrac{1240\ \text{km}}{696\ \frac{\text{km}}{\text{hr}} }\ =\ \dfrac{1240}{696}\ \text{hr}\)

\(\text{total time}\ =\ \text{time}_1+\text{time}_2 \ =\ \dfrac{620}{1032}\ \text{hr}\ +\ \dfrac{1240}{696}\ \text{hr}\ =\ \Big( \dfrac{620}{1032}\ +\ \dfrac{1240}{696}\Big)\ \text{hr}\ \approx\ 2.38\ \text{hr} \) _

Remember that raising a number to the power of  -1  "flips" it so the numerator becomes the denominator and vice versa.

Like this:          ( a / b )^-1   =   b / a          and          ( a )^-1   =   ( a / 1 )^-1   =   1 / a

\((i-i^{-1})^{-1}\\~\\ =\quad(i-\frac1i)^{-1}\)

                                     Get a common denominator between  \(i\)  and  \(\frac1i\)  by multiplying the first term by  \(\frac{i}{i}\)

\(=\quad(i\cdot\frac{i}{i}-\frac1i)^{-1}\\~\\ =\quad(\frac{i^2}{i}-\frac1i)^{-1}\\~\\ =\quad(\frac{i^2-1}{i})^{-1}\)

                                     Replace  \(i^2\)  with  -1

\(=\quad(\frac{-1-1}{i})^{-1}\\~\\ =\quad(\frac{-2}{i})^{-1}\\~\\ =\quad\frac{i}{-2}\\~\\ =\quad-\frac{1}{2}i\)

Since vertical angles are congruent, we can make the following equation:

(4x + 7)°   =   (7x - 35)°

                                           We can remove the degree signs and parenthesees from both sides of the equation.

4x + 7   =   7x - 35

                                           Subtract 4x from both sides, add  35  to both sides.

7 + 35   =   7x - 4x

                                           Combine like terms.

42   =   3x

                                           Divide both sides of the equation by  3

?   =   x

Alternate approach:

Since a  180°  angle forms a straight line, we can make the following equation:

(4x + 7)°  +  (10x - 23)°   =   180°

                                                       Remove the degree signs from every term, then we can drop the two

                                                       sets of parenthesees because there is only addition between them

4x + 7  +  10x - 23   =   180

                                                       Simplify the left side by combining like terms.

14x  -  16   =   180

                                                       Add  16  to both sides.

14x   =   196

                                                       Divide both sides of the equation by  14

x   =   ?

the length of PQ   +   the length of QR   =   the length of PR

(5x - 2)   +   (x + 12)   =   40

                                                   We can take off the parenthesees since there is only addition between them

5x - 2 + x + 12   =   40

                                                   Rearrange the terms

5x + x + 12 - 2   =   40

                                                   Combine like terms

6x + 10   =   40

                                Subtract  10  from both sides of the equation.

6x   =   30

                                Divide both sides of the equation by  6

x   =   5

Now that we know what  x  is, plug it into each expression to find  PQ  and  QR

PQ   =   5x - 2   =   ?

QR   =   x + 12   =   ?

x ≤ -2