CPhill

Press the "2nd" button on the calculator and then hit the "asin" button and then enter your value...then, hit enter or "="

145+3 < 15x     adding on the left, we have

148 < 15x      divide both sides by 15

148/15 < x   or

x > 148/15

Let (x, y) be the other point.

So we have, using the mid-point formula......

(3 + x) /2 = 3.5      and   (-4 + y)/2 = -3       multipling both equations by 2 on both sides

3 + x = 7               and     -4 + y  = -6         and solving for x and y, we have

-3        -3                         +4          +4

      x = 4               and            y = -2

So the point is   (4, -2)

1. If two numbers have an absolute value of 16, one must be 16 and the other must be -16. So, clearly, 16 is the number greater than 12.

2. All the possible values for the point would be the positive numbers > 10. If the original point was < -10, all the absoulte values of such points would be > 10.

Haley, I'll give you point for creativity !!!

If the question is (pi)^2.....it equals  about 9.87

Mmmmm...I still like Haley's answer better ......!!!!

(1/2)^4*(2)^3* (4)^2 =

(1/16)*(8)*(16) =

(1/2) * 16 =

8

OK....tracing this one through

We know one angle not between the sides (SSA)

A > 90

a < b

No triangle possible

I believe B is the correct choice

(2, -5) → (3, y)

To get the new x coordinate we multiplied 2 by a scale factor of 1.5

So -5 * 1.5 should = -7.5   instead of -2.5

Note that we have two triangles here. We know an angle in each triangle and the adjacent side to that angle in the two triangles - (12ft). And we want to know the opposite side in each triangle. So we can use the tangent function to find the opposite side in each case and add these together to get the total height.

So, the height is given by

h = 12tan(36) + 12tan(42) = about 19.5 ft.

For the first one, the remaining angle in the triangle is 17°

So, using the Law of Sines, we have

d/sin(38) = 10/sin(17)   multiply both sides by sin(38)

d =10*sin(38)/sin(17) = about 21 ft.

The other two problems are similar to this one I just worked (using The Law of Cosines):