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2 years ago

32b) What are the like terms in the expression 4y + 5 + 3y? *

Mathematics

1 answer:

pashok25 [27]2 years ago

3 0

Answer:

4y and 3y

Step-by-step explanation:

4y and 3y are like terms because they have y after the number. Every time when a math problem ask you to combine like terms, it'll be the ones that have the same variables or powers.

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Hey guys<br>im new here<br>please solve this for me with steps!<br>ill mark as the best answer

Vinil7 [7]

Answer:

The factors of $2(x+y)^2-9(x+y)-5$ is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=> $2(x+y)^2-9(x+y)-5$

To Find:

The factors of the polynomial =?

Solution:

Lets assume k = (x+y)

Then $2(x+y)^2-9(x+y)-5$ can be written as $2k^2-9k-5$

Now by using quadratic formula

k = $\frac{-b\pm\sqrt{(b^2-4ac}}{2a}$

where

a= 2

b= -9

c= -5

Substituting the values, we get

k = $\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}$

k = $\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}$

k = $\frac{-(-9) \pm \sqrt{(81+40)}}{4}$

k = $\frac{-(-9) \pm \sqrt{(121)}}{4}$

k = $\frac{-(-9) \pm 11}}{4}$

k= $\frac{ 9 \pm 11}{4}$

k = $\frac{20}{4}$ k = $\frac{-2}{4}$

$k_1 =5$ $k_2 = -\frac{1}{2}$

$2k^2-9k-5= 2(k-5)(k+\frac{1}{2})$

Solving the RHS we get

$\frac{2}{2}(k-5)(2k+1)$

$(k-5)(2k+1)$

Substituting k = x+y

$((x+y)-5)(2(x+y+1)$

$((x+y)-5)(2x+2y+1)$

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Suppose that y varies inversely with x. Use the information to find k, and then choose the equation of variation. X=2 when y=2

Vlad1618 [11]

Answer:

see explanation

Step-by-step explanation:

Given that y varies inversely as x then the equation relating them is

y = $\frac{k}{x}$ ← k is the constant of variation

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3 years ago

32b) What are the like terms in the expression 4y + 5 + 3y? *​

32b) What are the like terms in the expression 4y + 5 + 3y? *