Question

(7a+3,5b+21) • (2,1c+17d+11,5)и (2x+3,2y+7)•(14x+7y+2,1)​

Answer 1

Ответ:

(7a+3,5b+21) • (2,1c+17d+11,5) = 14,7ac + 119ad + 80,5а + 7,35bc + 59,5bd + 40,25b + 44,1c + 357d + 241,5

(2x+3,2y+7) • (14x+7y+2,1) = 28x^2 + 58,8xy + 102,2x + 22,4y^2 + 55,72y + 14,7

Answer 2

It looks like you have three expressions that you want to multiply together. Let's break down each multiplication separately:

1. (7a + 3, 5b + 21) • (2, 1c + 17d + 11.5):

To multiply two expressions, you'll need to apply the distributive property, which means multiplying each term in the first expression by each term in the second expression. Here's the result:

(7a + 3) • 2 = 14a + 6 (7a + 3) • (1c + 17d + 11.5) = 7ac + 119ad + 80.5a + 3c + 51d + 34.5

So, the result of the first multiplication is: (14a + 6, 7ac + 119ad + 80.5a + 3c + 51d + 34.5)

2. (2x + 3, 2y + 7) • (14x + 7y + 2, 1):

Using the same distributive property, we can multiply these expressions:

(2x + 3) • (14x + 7y + 2) = 28x^2 + 14xy + 4x + 42xy + 21y + 6 (2x + 3) • 1 = 2x + 3

So, the result of the second multiplication is: (28x^2 + 56xy + 4x + 21y + 9, 2x + 3)

Now, you have the results of the first two multiplications. If you want to multiply the third expression with the result of the second multiplication, you can use the distributive property again. Let's do that:

3. (28x^2 + 56xy + 4x + 21y + 9, 2x + 3) • (14x + 7y + 2, 1):

(28x^2 + 56xy + 4x + 21y + 9) • (14x + 7y + 2) = (28x^2)(14x) + (28x^2)(7y) + (28x^2)(2) + (56xy)(14x) + (56xy)(7y) + (56xy)(2) + (4x)(14x) + (4x)(7y) + (4x)(2) + (21y)(14x) + (21y)(7y) + (21y)(2) + (9)(14x) + (9)(7y) + (9)(2) = 392x^3 + 196x^2y + 56x^2 + 784xy^2 + 392xy + 112xy + 56x^2 + 28xy + 8x + 294xy^2 + 147y^3 + 42y^2 + 126x + 63y + 18

Now, the final result of the multiplication is: (392x^3 + 980x^2y + 112x^2 + 1478xy^2 + 519y^3 + 56x + 63y + 18)