Answer:
you can do a diagram representing this product by drawing a box with six squares and then putting three lines for each box and divide 3÷3 and you will get 5
You have to write the equation for a line that crosses the point (-4, -7) and is perpendicular to the line

When you have to determine a line that is perpendicular to a known line, you have to keep in mind that the slope of the perpendicular line will be the negative inverse of the first one.
If for exampla you have two lines, the first one being:

And the second one, that is perpedicular to the one above:

The slope of the second one is the negative inverse of the first one:

The slope of the given line y=-7/4+4 is m=-7/4
So the slope of the perpendicular line has to ve the inverse negative of -7/4

Considering it has to pass through the point (-4,-7) and that we already determined its slope, you can unse the point slope formula to determine the equation of the perpendicular line:

replace with the coordinates of the point and the slope and calculate:

Subtract 7 to both sides of the equation to write it in slope-intercept form:

Now you can graph both lines
Step-by-step explanation:
x² + 3x + 2
x² + x + 2x + 2 because 1+2=3 and 2×1=2
By taking out common terms, we get
x(x+1)+2(x+1)
(x+1)(x+2)
Answer:
The answer is D) 12
Step-by-step explanation:
Multiple 4 by 3
The third term of the expansion is 6a^2b^2
<h3>How to determine the third term of the
expansion?</h3>
The binomial term is given as
(a - b)^4
The r-th term of the expansion is calculated using
r-th term = C(n, r - 1) * x^(n - r + 1) * y^(r - 1)
So, we have
3rd term = C(4, 3 - 1) * (a)^(4 - 3 + 1) * (-b)^(3-1)
Evaluate the sum and the difference
3rd term = C(4, 2) * (a)^2 * (-b)^2
Evaluate the exponents
3rd term = C(4, 2) * a^2b^2
Evaluate the combination expression
3rd term = 6 * a^2b^2
Evaluate the product
3rd term = 6a^2b^2
Hence, the third term of the expansion is 6a^2b^2
Read more about binomial expansion at
brainly.com/question/13602562
#SPJ1