Answer:
The 100,000,000
Step-by-step explanation:
(3 cos x-4 sin x)+(3sin x+4 cos x)=5
(3cos x+4cos x)+(-4sin x+3 sin x)=5
7 cos x-sin x=5
7cos x=5+sin x
(7 cos x)²=(5+sinx)²
49 cos²x=25+10 sinx+sin²x
49(1-sin²x)=25+10 sinx+sin²x
49-49sin²x=25+10sinx+sin²x
50 sin² x+10sinx-24=0
Sin x=[-10⁺₋√(100+4800)]/100=(-10⁺₋70)/100
We have two possible solutions:
sinx =(-10-70)/100=-0.8
x=sin⁻¹ (-0.8)=-53.13º (360º-53.13º=306.87)
sinx=(-10+70)/100=0.6
x=sin⁻¹ 0.6=36.87º
The solutions when 0≤x≤360º are: 36.87º and 306.87º.
9514 1404 393
Answer:
y = -(x+1)^2 +3
Step-by-step explanation:
Translating f(x) left by 1 unit replaces x with x+1.
Translating f(x) up by 3 units replaces f(x) with f(x)+3.
Reflecting f(x) over the x-axis replaces f(x) with -f(x).
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When y = x^2 is reflected over the x-axis, it becomes ...
y = -x^2
When y = -x^2 is translated 1 unit left, it becomes ...
y = -(x +1)^2
When y = -(x+1)^2 is translated 3 units up, it becomes ...
y = -(x +1)^2 +3
Answer:
- zeros: x = -3, -1, +2.
- end behavior: as x approaches -∞, f(x) approaches -∞.
Step-by-step explanation:
I like to use a graphing calculator for finding the zeros of higher order polynomials. The attachment shows them to be at x = -3, -1, +2.
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The zeros can also be found by trial and error, trying the choices offered by the rational root theorem: ±1, ±2, ±3, ±6. It is easiest to try ±1. Doing so shows that -1 is a root, and the residual quadratic is ...
x² +x -6
which factors as (x -2)(x +3), so telling you the remaining roots are -3 and +2.
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For any odd-degree polynomial with a positive leading coefficient, the sign of the function will match the sign of x when the magnitude of x gets large. Thus as x approaches negative infinity, so does f(x).
Subtract 363,321 from 778,871. Which gives you 415,550 more geese than crows :)