Full question:
For each multiplication expression, sketch an area model. Label the dimensions and the area of each part. Then write an equation showing that the area as a product equals the area as a sum. a. (x+1)(x+2), b. 3(2x+5), c. (2x-3)(x+2), d. (x-1)(y-1), e. -2y(y+3), f. (-x+1)(3x+y-4)
Answer and explanation:
a. (x+1)(x+2)= x×x+x×2+1×x+1×2
The dimensions (length and width) is x+1 and x+2
b. 3(2x+5) = 3×2x+3×5
The dimensions is 3 and 2x+5
c. (2x-3)(x+2)= 2x×x+2x×2-3×x-3×+2
The dimensions are 2x-3 and x+2
d. (x-1)(y-1)= x×y+x×-1-1×y-1×-1
Dimensions are x-1 and y-1
e. -2y(y+3)= -2y×y-2y×3
Dimensions are -2y and y+3
f. (-x+1)(3x+y-4)= -x×3x-x×y-x×-4+1×3x+1×y+1×-4
Dimensions are -x+1 and 3x+y-4
Answer:
2.89, rounded to the nearest hundredth
Step-by-step explanation:
Given that GPA is weighted by credits, we must first multiply each grade by its credit amount and sum those up to weigh the credits. Then, we divide by the total amount of credits to get the GPA per credit.
So, we start with math,
3.7 *5 + 1.8 *3 + 2.8 * 5 + 2.8 * 4 = 49.1 as the total GPA weighted per credit.
Then, to find the average per credit, we divide by the total amount of credits, which is 5 + 3 + 5 + 4 = 17.
Our answer is 49.1/17 = 2.89, rounded to the nearest hundredth
Answer:
15
Step-by-step explanation:
Add the total of miles Marisa is going to drive; 200 + 220 = 420
Next, divide the total miles by the average miles Marisa can travel on one gallon of gas; 420 / 28 = 15
(Hope this helps!)
Answer:
a. 6
b. 9
Step-by-step explanation:
a. The product modulo 7 can be found from the product of the individual numbers modulo 7:
(88·95·36·702) mod 7 = (88 mod 7)·(95 mod 7)·(36 mod 7)·(703 mod 7) mod 7
= (4·4·1·3) mod 7 = 48 mod 7 = 6
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b. Powers of 4 mod 11 repeat with period 5:
4 mod 11 = 4
4^2 mod 11 = 5
4^3 mod 11 = 9
4^4 mod 11 = 3
4^5 mod 11 = 1
So, 4^83 mod 11 = 4^3 mod 11 = 9
The first reaction represented is a decomposition reaction due to the fact the bonded pair are being split apart.
The second equation is a single displacement because the B and C have switched places, which is the only change making it a single displacement. If there had been another bond and A had also moved, it would be double.