Answer:
n = 59
Step-by-step explanation:
I find it easiest to work problems of this kind using a graphing calculator. That way, extraneous solutions can be avoided. It seems to work well to rewrite the problem, so you're looking for a value of n that makes the result zero. Here, that would mean you want ...
... f(n) = √(n+5) -√(n-10) -1
_____
The solution by hand involves eliminating the root symbols. You do that by squaring the equation:
... n +5 -2√((n+5)(n-10)) + n -10 = 1
Now, we isolate the remaining root and square again.
... 2n -6 = 2√((n+5)(n-10)) . . . collect terms, add 2√( ) -1
... n -3 = √(n²-5n-50) . . . . . . . divide by 2
... n² -6n +9 = n² -5n -50 . . . . square both sides
... 59 = n . . . . . . . . . . . . . . . . . add 50 +6n -n²
Answer:
49/1
Step-by-step explanation:
The point is on (1,49)
520, maybe 39 + 65 is 104 104 x 5 is 520 not sure if tax is included in the equation, but if not 520 should be correct
The x, which is zero in this case will always be the number that goes on the inside of the equation, so far you have y=(x+0)^2 or y=x^2.
The y, which is negative will go on the end of the equation. So you get y=x^2-5
The equation is y=x^2-5 or C if it is multiple choice
Note that a squared pyramid has a square base & 4 equal triangles.
To find the lateral the lateral area you calculate the area of the 4 equal triangles and to find the surface area (total Area) you add the area of the base:
Area of each triangle = side (5) x slant (9) and you divide by 2
==>Aera of 1 triangle = (9x5)/2 = 45/2 & for the 4 triangles
Lateral area = (45/2) x 4 = 90 in²
Now the base area (square) = 5 x 5= 25 in²
so the surface area = 90+25 = 115 in² (answer a)
