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3 years ago

6(x+xy)-5(y+2xy) how do I solve this problem?

Mathematics

2 answers:

Anastasy [175]3 years ago

5 0

Answer:

To solve this problem you would have to distribute the 5 and the 6. Once that is completed you would find the matching pairs and add or subtract them.

The answer is: 6x-4xy-5y

Step-by-step explanation:

shepuryov [24]3 years ago

4 0

$6(x+xy)-5(y+2xy)=6x+6xy-(5y+10xy) \\6x+6xy-5y-10xy \\\boxed{6x-5y-4xy}$

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Select the correct comparison.

Gnesinka [82]

Answer:

B. The typical value is greater in set A. The spread is greater in set B.

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2 years ago

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What is the slope of the line below

Fofino [41]

Answer:

C. $\frac{1}{2}$

Step-by-step explanation:

You can use the formula to find the slope: $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

(-1.5, 1.5) & (1.5, 0)

$\frac{0-(-1.5)}{1.5-(-1.5)} =\\\\\frac{0+1.5}{1.5+1.5} =\\\\\frac{1.5}{3} =\\\\\frac{1}{2}$

The slope is $\frac{1}{2}$

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3 years ago

Express 2x+1/(x-2)(x²+1) as a partial fraction.

Oduvanchick [21]

Answer:

Partial fraction = 1/(x-2) - x/(x^2+1)

Step-by-step explanation:

Question:

Express 2x+1/(x-2)(x²+1) as a partial fraction.

Note: it will be assumed that there was a typo in the interpretation of parentheses to mean

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

Let

(2x+1) / ( (x-2)(x^2+1) ) = A/(x-2) + (Bx+C)/(x^2+1) .........................(0)

(2x+1) / ( (x-2)(x^2+1) ) = (A(x^2+1)+(Bx+C)(x-2)) / ( (x-2)(B/(x^2+1) )

(2x+1) / ( (x-2)(x^2+1) ) = (Ax^2+A+Bx^2+(C-2B)x-2C) / ( (x-2)(B/(x^2+1) )

(2x+1) / ( (x-2)(x^2+1) ) = ( (A+B)x^2+(C-2B)x+A-2C ) / ( (x-2)(B/(x^2+1) )

Match numerators

2x+1 = (A+B)x^2+(C-2B)x+A-2C

Match coefficients,

A+B = 0 ..................(1)

-2B+C = 2 .................(2)

A-2C = 1 ...................(3)

Solve for A, B and C

Substitute A from (1) in (3)

-B - 2C =1

transpose and solve for B

B = -2C-1 ....................(4)

Substitue B from (4) in (2)

-2(-2C-1) + C = 2

simplify

5C = 2-2 = 0

C=0 ..........................(5)

substitute (5) in (4)

B = -2C-1 = -1 ...............(6)

Substitue (6) in (1)

A+(-1) = 0

A=1 .............................(7)

Using values from (7), (6) and (5) to substitute in (0)

we get

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

as the required partial fraction

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3 years ago

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225 to 270 I need help please help me

vagabundo [1.1K]

Wheres the picture or the problem????

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3 years ago

The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there a

STatiana [176]

Answer:

24761 mosquitoes

5.4 days

Step-by-step explanation:

Let the equation that shows the population of the mosquitoes after x days,

$A=P(1+r)^x$

Where,

P = Initial population,

r = rate of increasing per day,

Here, A = 1000, when x = 0,

$1000=P(1+r)^0\implies P = 1000$

A = 1900 when x = 1,

$1900 = P(1+r)^1\implies 1900 = 1000(1+r)\implies 1.9 = 1 + r\implies r = 0.9$

Thus, the required function that represents the population after x days,

$A=1900(1+0.9)^x=1900(1.9)^x$

If x = 4,

The number of mosquito after 4 days,

$A=1900(1.9)^4 = 24760.99\approx 24761$

If A = 60,000,

$60000 = 1900(1.9)^x$

$\frac{600}{19}=1.9^x$

Taking log both sides,

$\log(\frac{600}{19})=x\log 1.9$

$\implies x = \frac{\log(\frac{600}{19})}{\log 1.9}=5.378\approx 5.4$

Thus, there will 60,000 mosquitoes after 5 days.

6 0

3 years ago

6(x+xy)-5(y+2xy) how do I solve this problem?​

6(x+xy)-5(y+2xy) how do I solve this problem?