To find the product of (4x-5y)^2,
we can rewrite the problem as:
(4x-5y)(4x-5y) (two times because it is squared)
Now, time to use that old method we learned in middle school:
FOIL. (Firsts, Outers, Inners, and Lasts)
FOIL can help us greatly in this scenario.
Let's start by multiplying the 'Firsts' together:
4x * 4x = <em>16x^2</em>
Now, lets to the 'Outers':
4x * -5y = <em>-20xy</em>
Next, we can multiply the 'Inners':
-5y * 4x = <em>-20xy</em>
Finally, let's do the 'Lasts':
-5y * -5y = <em>25y</em>^2
Now, we can take the products of these equations from FOIL and combine like terms. We have: 16x^2, -20xy, -20xy, and 25y^2.
-20xy and -20xy make -40xy.
The final equation (product of (4x-5y)^2) is:
16x^2 - 40xy + 25y^2
Hope I helped! If any of my math is wrong, please report and let me know!
Have a good one.
Answer: b. -2x+3y=12
Step-by-step explanation: the y intercept is 4, and the line intercepts the x- axis at -6. Therefore, the slope of the line is 4/6=2/3. The format for a linear equation is y=(slope)x+y intercept. So the equation would be y=2/3x+4. You can see that answer b simplifies to that. Add 2x from both sides of the equation and you get 3y=2x+12. Divide both sides of the equation by 3 and you get y=2/3x+4
(6 1/3) / (5/6) = (19/3) / (5/6) = 19/3 * 6/5 = 114/15 = 38/5
3 - 7^2 + 3 * 4^2
3 - 49 + 3 * 16
3 - 49 + 48
- 46 + 48
2
(7b - 2) / (-a + 1)....a = -2, b = 3, c = -1/3
where is he c in this equation ? Because if I just use a and b, my answer is not an answer choice
(7b - 10) / (a - 1)...a = -1, b = 5, c = -2/3
same as before...where is the c in this equation
V = (pi) r^2* h
V = (pi)(5^2)(7)
V = (pi)(25)(7)
V = 175(pi)
The reciprocal of a given fraction is obtained by flipping the elements of the fraction. This means that the current numerator becomes the denominator of the reciprocal and the current denominator becomes the new numerator. Thus, for the given fraction, the reciprocal is 9/7.
