hectictar

Here's a rough drawing of the triangle:

Since the sum of the angle measures in a triangle is  180°...

m∠BCA  =  180° - 54° - 52°   =   74°

Now look at quadrilateral CDHE.

Since the sum of the angle measures in a quadrilateral is  360°...

m∠DHE  =  360° - 74° - 90° - 90°  =  106°

And since  ∠DHE  and  ∠AHB  are vertical angles, they have the same measure.

m∠AHB  =  m∠DHE  =  106°

Alice is one the line  x = 4  ,  so the  x-coordinate of whatever point Alice is on is  4 .

Let's call the point that Alice is located  (4, y)  .

By the distance forumula...

the distance between  (4, y)  and  (2, 3)   =   \(\sqrt{(4-2)^2+(y-3)^2}\)

The problem tells us that the distance between  (4, y)  and  (2, 3)  is  √40 ,  so...

\(\sqrt{(4-2)^2+(y-3)^2}=\sqrt{40}\)

                                                          Now let's solve this equation for  y .  First square both sides.

\((4-2)^2+(y-3)^2=40\)

                                                          Simplify  \((4-2)^2\)  to  \(4\) .

\(4+(y-3)^2=40\)

                                                          Subtract  4  from both sides of the equation.

\((y-3)^2=36\)

                                                          Take the ± square root of both sides.

\(y-3=\pm\sqrt{36}\)

\(y-3=\pm6\)

                                                          Add  3  to both sides.

\(y=3\pm6\)

\(\begin{array}{ccc} y=3+6&\qquad\text{or}\qquad &y=3-6\\~\\ y=9&\text{or}&y=-3 \end{array}\)

The possible y-coordinates of the point Alice at are  9  and  -3 .

The sum of the possible y-coordinates of the point Alice is at   =   9 + -3   =   6

3)

Let the number of nickels in the bag be  n .

Let the number of dimes in the bag be  d .

There are 3 times as many dimes as nickels. In other words...

The number of dimes is 3 times the number of nickels.

d  =  3n

He has $36.05 in his bag. So he has a total of 3605 cents in his bag.

A nickel is worth 5 cents, and a dime is worth 10 cents. So...

5n + 10d  =  3605

                                    Substitute  3n  in for  d .

5n + 10(3n)  =  3605

                                    Multiply  10  and  3  to get  30 .

5n + 30n  =  3605

                                    Combine like terms.

35n  =  3605

                                    Divide both sides of the equation by  35 .

n  =  103

2)

Let the amount of money that I have be  a .

Let the amount of money that my sister has be  b .

If I give my sister 17 dollars, then we will have the same amount of money.

a - 17  =  b + 17

                               Add  17  to both sides of the equation to solve for  a .

a  =  b + 34

If instead she gives me 23 dollars, then I'll have three times as much money as she will have.

a + 23  =  3(b - 23)

                                         Since  a  =  b + 34 ,  we can substitute  b + 34  in for  a .

b + 34 + 23  =  3(b - 23)

                                         Simplify both sides.

b + 57  =  3b - 69

                                         Add  69  to both sides.

b + 126  =  3b

                                         Subtract  b  from both sides.

126  =  2b

                                         Divide both sides by  2 .

63  =  b

The sister currently has  63  dollars. And you currently have  63 + 34 dollars, which is 97 dollars. You can try to imagine actually playing out the scenarios, knowing these amounts of money to start with, to help understand. 

I think all your answers are correct! 👍

I got the same answer for each one:

k(m)  =  m² - 7m - 18

k(m)  =  (m - 9)(m + 2)

The zeros are the values of  m  that make  k(m)  equal zero, so the zeros are  9  and  -2 .

C)  One positive zero and one negative zero, where the positive zero has the larger absolute value

f(x)  =  2x² - 11x + 12

f(x)  =  2x² - 8x - 3x + 12

f(x)  =  2x(x - 4) - 3(x - 4)

f(x)  =  (x - 4)(2x - 3)

The zeros are  4  and  3/2  .

D)  Two positive zeros

s(r)  =  5r² + 49r - 10

s(r)  =  5r² + 50r - 1r - 10

s(r)  =  5r(r + 10) - 1(r + 10)

s(r)  =  (r + 10)(5r - 1)

The zeros are  -10  and  1/5 .

F)  One positive zero and one negative zero, where the negative zero has the larger absolute value

z(w)  =  w² + 7w + 6

z(w)  =  (w + 6)(w + 1)

The zeros are  -6  and  -1 .

E)  Two negative zeros

For the third question...

∠A  and  ∠B  must be acute angles because the third angle is 90°, and the sum of the angle measures in a triangle is 180°, and each angle has to have a measure greater than 0°. (If either  ∠A  or  ∠B  were  90°  then two of the sides would be parallel, and they could never meet.)

First let's find the sine and cosine of angle A .

sin A  =  opposite / hypotenuse

sin A  =  12 / 20

sin A  =  3 / 5

cos A  =  adjacent / hypotenuse

cos A  =  16 / 20

cos A  =  4 / 5

Next let's find the sine and cosine of angle B .

sin B  =  opposite / hypotenuse

sin B  =  16 / 20

sin B  =  4 / 5

cos B  =  adjacent / hypotenuse

cos B  =  12 / 20

cos B  =  3 / 5

S  =  2r³t

                       To solve this equation for  t ,  first let's divide both sides by  2 .

\(\frac{S}{2}\)  =  r³t

                       Next divide both sides by  r³ .

\(\frac{S}{2r^3}\)  =  t

t  =  \(\frac{S}{2r^3}\)

---------------------------

4x² - 376  =  24

                                 To solve this equation for  x , first let's add  376  to both sides.

4x²  =  24 + 376

4x²  =  400

                                 Then divide both sides by  4 .

x²  =  400/4

x²  =  100

                                 Then take the ± square root of both sides.

x  =  ±√100

x  =  ± 10

This means there are two values of  x  that make the equation true,

and those two values are 10  and  -10 .

Maybe your teacher wants you to factor the given expression as a difference of squares? Like this....

\({\color{RedOrange}a}^2-{\color{JungleGreen}b}^2= ({\color{RedOrange}a}+{\color{JungleGreen}b})({\color{RedOrange}a}-{\color{JungleGreen}b}) \)          so.....

\({\color{RedOrange}(4j-2)}^2-{\color{JungleGreen}(2+4j)}^2\,=\, ({\color{RedOrange}(4j-2)}+{\color{JungleGreen}(2+4j)})({\color{RedOrange}(4j-2)}-{\color{JungleGreen}(2+4j)}) \\~\\ \phantom{(4j-2)^2-(2+4j)^2}\,=\,((4j-2)+(2+4j))((4j-2)-(2+4j))\)

Simplifying further...

\( \phantom{(4j-2)^2-(2+4j)^2}\,=\,(4j-2+2+4j)(4j-2-2-4j)\\~\\ \phantom{(4j-2)^2-(2+4j)^2}\,=\,(4j+4j)(-2-2)\\~\\ \phantom{(4j-2)^2-(2+4j)^2}\,=\,(8j)(-4)\\~\\ \phantom{(4j-2)^2-(2+4j)^2}\,=\,-32j\)

If you want to factor  -32j  , then you want to get it into its prime factors.

( For example, if you chose to factor it into  -1 * 4 * 8 * j  then you could still factor out more,

  because  4 = 2 * 2  and  8 =  4 * 2    )

-32j   =   -1 * 2 * 2 * 2 * 2 * 2   =   -2⁵j    in its prime factored form. 

We are given that   AB  =  3AD  =  6BC   so...

3AD  =  6BC

                          Divide both sides of the equation by  3 .

AD  =  2BC

And we can see that....

CD  =  AB - AD - BC

                                         Substitute  6BC  in for  AB  and substitute  2BC  in for  AD .

CD  =  6BC - 2BC - BC

                                         Combine like terms.  (  6x - 2x - x  =  3x    so    6BC - 2BC - BC  =  3BC  )

CD  =  3BC

probability that a random point lies on CD  =  \(\frac{\text{CD}}{\text{AB}}\)

                                                                                   Substitute  3BC  in for  CD  and  6BC  in for  AB .

probability that a random point lies on CD  =  \(\frac{3\text{BC}}{6\text{BC}}\)

                                                                                   Reduce fraction by the factor  BC .

probability that a random point lies on CD  =  \(\frac36\)

                                                                                   Reduce fraction by  3 .

probability that a random point lies on CD  =  \(\frac12\)

If there exists an integer  a  such that  f(a) = b ,   then  f(b)  is defined and

f(b) = 3b + 1  if  b  is odd      and      f(b) = b/2  if  b  is even .

We must have at least these defined values of the function:

f(14) = 7

f(7) = 22

f(22) = 11

f(11) = 34

f(34) = 17

f(17) = 52

f(52) = 26

f(26) = 13

f(13) = 40

f(40) = 20

f(20) = 10

f(10) = 5

f(5) = 16

f(16) = 8

f(8) = 4

f(4) = 2

f(2) = 1

f(1) = 2

There must be at least 18  integers in the domain of  f .