hectictar

1.

By the Law of Cosines:

\(a^2 = c^2 + b^2 - 2cb \cos A\\~\\ (\overline{\text{BC}})^2 \,=\,(7\sqrt2)^2+(6)^2-2(7\sqrt2)(6)\cos(45°)\\~\\ (\overline{\text{BC}})^2 \,=\,98+36-(84\sqrt2)(\frac{\sqrt2}{2})\\~\\ (\overline{\text{BC}})^2 \,=\,98+36-84\\~\\ (\overline{\text{BC}})^2 \,=\,50\\~\\ \overline{\text{BC}} \,=\,\sqrt{50}\\~\\ \overline{\text{BC}} \,=\,5\sqrt{2}\)

1.  all whole numbers

2.

Below are all possible outcomes:

2, 2       4, 2       7, 2       9, 2

2, 4       4, 4       7, 4       9, 4

2, 7       4, 7       7, 7       9, 7

2, 9       4, 9       7, 9       9, 9

The ones colored blue are the ones which have a product less than 30 and contain at least one odd number.

There  are  6  of them.

3.

y  =  x + 6

y  =  x² + 8x + 16                Substitute  x + 6  in for  y  in this equation.

x + 6  =  x² + 8x + 16

                                            Subtract  x  and subtract  6  from both sides.

0  =  x² + 7x + 10

                                            Factor the right side.

0  =  (x + 2)(x + 5)

                                            Set each factor equal to zero and solve for  x .

x + 2 = 0     or     x + 5 = 0

 x  =  -2       or       x  =  -5

When  x  =  -2 ,     y  =  x + 6  =  -2 + 6  =  4       So  (-2, 4)  is a solution.

When  x  =  -5 ,     y  =  x + 6  =  -5 + 6  =  1       So  (-5, 1)  is a solution.

A.

volume of rectangular prism  =  length * width * height

volume of rectangular prism  =  20 cm * 9 cm * 8 cm

volume of rectangular prism  =  1440 cm²

B.

volume of circular cylinder  =  area of circular base * height

volume of circular cylinder  =  π * radius² * height

radius  =  diameter / 2  =  10/2 cm  =  5 cm

volume of circular cylinder  =  π * (5 cm)² * 15 cm

volume of circular cylinder  =  π * 25 * 15 cm³

volume of circular cylinder  =  375π  cm³

volume of circular cylinder  ≈  1178  cm³        (That is to the nearest cubic centimeter.)

C.

volume of remaining wax   =   volume of rectangular prism - volume of circular cylinder

volume of remaining wax   =   1440 cm³  -  375π  cm³

volume of remaining wax   =   ( 1440 - 375π ) cm³

volume of cube  =  volume of remaining wax  =  ( 1440 - 375π ) cm³

volume of cube  =  side * side * side  =  ( side )³

                                                        Equate both representations of volume of cube.

( side )³  =  ( 1440 - 375π ) cm³

                                                        Take the cube root of both sides of the equation.

side  =  \(\sqrt[3]{1440-375\pi}\)  cm

                                                        Plug  \(\sqrt[3]{1440-375\pi}\)  into a calculator.

side  ≈  6 cm

This is what I found....

Let's label the figure like this:

A line from the center of a circle to a point of tangency is perpendicular to its tangent line, so  CD  is perpendicular to  AB .

And so  △CDA  is a right triangle  and  △CDB  is a right triangle.

By the Pythagorean theorem:

\(r^2+s^2\,=\,16^2\)

and

\(r^2+t^2\,=\,(\frac{15}{2})^2\)

and

\(16^2+(\frac{15}{2})^2=(s+t)^2\)

Now we have 3 equations and 3 unknowns.

Now I am really confused....because when I used WolframAlpha to solve the system of equations, it found that

\(r\,=\,\frac{240}{\sqrt{1249}}\,=\,\frac{240\sqrt{1249}}{1249}\,\approx\,6.791\)

A few things I noticed:

Other than that technical stuff I think it's really well written and easy to follow 👍

The volume of the prism is three times the volume of the pyramid.

The formula for the volume of a pyramid with height  h  and base area  B  is

V  =  \(\frac13\)Bh

△ABC ~ △DEF   means   "triangle ABC is similar to triangle DEF."

If you take a triangle and multiply each of its sides by some number, you will get a similar triangle.

For example:

If you take  △ABC  and multiply each of its sides by 2, you will get  △DEF.  So  △ABC ~ △DEF

( Also note that the order of the letters in the triangle name matters.

  For example, in this case it is NOT true that  △ABC ~ △FDE )

3.)

cos θ  =  x / r    where  x  is the  x-coordinate of  C  and  r  is the radius.

cos θ  =  x / 26

Let  x  and  y  be the coordinates of point  C. We know...

x² + y²  =  26²

x² + 24²  =  26²

x²  =  26² - 24²

x  =  ±√[ 26² - 24² ]

x  =  ±√[ 100 ]

x  =  ± 10

Point C  is in Quadrant II, so the  x-coordinate of  point C  must be negative.

x  =  -10

cos θ  =  x / 26

cos θ  =  -10 / 26

cos θ  =  -5 / 13

The dolphins at the Scott Aquarium are fed 1 1/2 buckets of fish each day. The sea otters are fed

4/5 as much fish as the dolphins. How many buckets of fish are the sea otters fed each day?

_________________________________________________________________________

The number of buckets of fish the dolphins are fed  =  \(1+\frac12\)

The sea otters are fed  \(\frac45\)  as much fish as the dolphins. In other words...

the number of buckets of fish the sea otters are fed  =  \(\frac45\) the number of buckets of fish the dolphins are fed

the number of buckets of fish the sea otters are fed  =  \(\frac45\) * (\(1+\frac12\))

the number of buckets of fish the sea otters are fed  =  \(\frac45\) * \((\frac32)\)

the number of buckets of fish the sea otters are fed  =  \(\frac{12}{10}\)

the number of buckets of fish the sea otters are fed  =  \(\frac65\)