No, a cubic equation can not have three complex roots. This is because it turns twice and one end goes to positive infinity and one end goes to negative infinity. Thus, one of these MUST cross the x-axis at some point, meaning y = 0 and a real root exists.
Yes, a cubic equation can have three real roots if it cuts the x-axis three times.
Answer: the answer is -16x+7y+4z
Step-by-step explanation:
You have an angle of elevation of 3 degrees and you're 2000 ft from base of 30 story building.
<span>Draw a picture of this. Then tan(3) = ht of bldg/2000 </span>
<span>I get a height of 104.82 ft rounded to 2 dp. </span>
<span>5. Ok. use the Pythagorean Theorem here to find the hypotenuse of the right triangle </span>
<span>hypt = sqrt(50^2 + 9^2) </span>
<span>Now sine of the angle of elevation is 50/hypt. = 0.984 or 0.98 to 2 dp.</span>
The original width would be 19 and the original length would be 32.
Let w be the width. Then 2w-6 would be the length. However, after cutting a 3-inch square from each corner, both the width and length left over to fold into a box would be 6 inches smaller; thus the dimensions would be w-6 and 2w-6-6 or 2w-12.
Since the section cut out is 3 inches long, 3 will be the height of the box.
Volume is found by multiplying the length, width and height of the box; thus we have:
1014=(w-6)(2w-12)(3)
We multiply the binomials and have:
1014 = [w*2w-12*w-6*2w-6(-12)](3)
1014 = (2w²-12w-12w+72)(3)
1014 = (2w²-24w+72)(3)
1014 = 6w² - 72w + 216
When solving a quadratic equation, we want it set equal to 0. Subtract 1014 from each side:
1014-1014 = 6w² - 72w + 216 - 1014
0 = 6w² - 72w - 798
We will use the quadratic formula to solve this:

Since we cannot have a negative number for a measurement, 19 has to be the width; then 2(19)-6 = 32 would be the length.
Answer:
it would be a close to $20