The middle number is 0 and the last is -81Factoring means we want something like(b+_)(b+_)We need two numbers that add together to get and Multiply to get -81which are 9 and -9:<span><span>9+-9 = 0 and </span><span>9*-9 = -81</span></span>
Fill in the blanks in
(b+_)(b+_)
with 9 and -9 to get...<span><span>(<span>b+9</span>)</span><span>(<span>b−9</span>)</span></span>
The answer is : (b+9)(b−9)
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Answer:
It is again the fourth answer VerifiedExpert
Step-by-step explanation:
If you graph of sub in values, you will see no answer dips below (-7,0).
Hope it helps.
Answer:
5.0 would be your answer.
Have a great day/night!
Answer:
y= 3x -4
Step-by-step explanation:
The equation of a line can be written in the form of y=mx +c, where m is the slope and c is the y-intercept. This is also known as the slope-intercept form.
Since the given equation is in the slope-intercept form, we can identify its slope from the coefficient of x.
Slope= -⅓
The product of the slopes of perpendicular lines is -1.
Slope of perpendicular line
= 3
Thus, the equation of the perpendicular line is given by:
y= 3x +c
Substitute a pair of coordinates that the line passes through to find the value of c.
When x= 3, y= 5,
5= 3(3) +c
5= 9 +c
<em>Minus 9 on both sides:</em>
c= 5 -9
c= -4
Hence, the equation of the perpendicular line is y= 3x -4.
Additional:
For more questions on equation of perpendicular lines, do check out the following!
Answer:
<h2>50.8°</h2>
Step-by-step explanation:
First, we need to use the Pythagorean Theorem to find the third side. We will then use the law of cosines to find the angle.
2√15² = 2√6² + b²
60 = 24 + b²
Subtract 24
36 = b²
b = 6
Now, that we have all three sides, we will use the law of cosines to figure out the angle.
c² = a² + b² - 2abCosC°
Substitute the values in.
6² = 2√15² + 2√6² - 2(2√15)(2√6)CosC°
36 = 60 + 24 - 2(37.947331922)CosC°
36 = 60 + 24 - 75.894663844CosC°
36 = 84 - 75.894663844CosC°
Subtract 84
-48 = -75.9CosC°
0.63241106719 = CosC°
Arccos (0.63241106719) = 50.77176844°
The largest acute angle is approximately 50.8°
