Mathematics, 19.03.2020 00:22 bbqchicken243

Each shape is 1 whole for number 13a-13e, choose yes or no to show whether the names the parts that are shaded

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Answer from: Quest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer: (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

Answer from: Quest

i dont know the answer but here's a picture of a dog

Answer from: Quest

$\boxed{\bold{2240a^4u^{18}w^3x^4}}$

explanation:

remove parenthesis: $\bold{\left(a\right)=a,\: -\left(-a\right)=a}$

= $\bold{4u^9w^1\cdot \: 7x^4\cdot \: 4a^3w^2\cdot \: 8\cdot \frac{1}{2}a^1\cdot \: 5u^9}$

apply rule $\bold{a^1=a}$

= $\bold{4\cdot \: 7\cdot \: 4\cdot \: 8\cdot \: 5\cdot \frac{1}{2}a^3au^9u^9w^2wx^4}$

multiply fractions: $\bold{\: a\cdot \frac{b}{c}=\frac{a\: \cdot \: b}{c}}$

= $\bold{\frac{1\cdot \: 4u^9w\cdot \: 7x^4\cdot \: 4a^3w^2\cdot \: 8a\cdot \: 5u^9}{2}}$

$\bold{1\cdot \: 4u^9w\cdot \: 7x^4\cdot \: 4a^3w^2\cdot \: 8a\cdot \: 5u^9: \ 4480x^4w^3a^4u^{18}}$

= $\bold{\frac{4480a^4u^{18}w^3x^4}{2}}$

divide: $\bold{\frac{4480}{2}=2240}$

= $\bold{2240a^4u^{18}w^3x^4}$

mordancy.

Answer from: Quest

yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.

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