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2 years ago

If 3x + 9y = 21 , find the value of 4(x + 3y)

Mathematics

1 answer:

mr Goodwill [35]2 years ago

7 0

Answer:

Step-by-step explanation:

The method that should be used is substitution:

Do this by taking 3x+9y=21 and transforming it to be $y=-\frac{1}{2} x+\frac{7}{3}$

Once you have this, substitute the value of y that we just found into 3x+9y=21 to find the value of x: $3x+9(-\frac{1}{2} +\frac{7}{3} ) =21$

Solve for x. You should get 1.5

Once you have this, Plug in 1.5 for the value of x into the y= equation that we found in the beginning: $y=-\frac{1}{2} (1.5)+\frac{7}{3}$

Solve for y. You should get 1.583 (19/12)

Plug in the values of x and y that we found into the last equation to find its value: 4(1.5+3(1.583))

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7 1/3 + 5 5/6 sum up as a fraction please 100 points

beks73 [17]

Answer:

The answer is 13 1/6

Step-by-step explanation:

The answer is 18 because 7 1/3 turns into 7 2/6 and 7 * 6 is 42 and 42 plus 2 is 44. Next You turn 5 5/6 into a improper fraction which turns into 35/6 and 35 plus 44 is 79/6 and 79/6 simplified is 13 1/6

3 0

2 years ago

Fill in the boxes please

IRINA_888 [86]

Answer:

x=12, DO=45O, G=45, DG=90

Step-by-step explanation:

Mid point means both lenghts are equal

4x-3=2x+21

2x=24

X=12

-----------------

1)4(12)-3=48-3=45

2)2(12)+21=24+21=45

-----------------------------

DO=45

OG=45

DG=90

4 0

3 years ago

Read 2 more answers

Text wants to make a lawn in front of his house. He buys enough sod to make a 10 ft.by 30 ft. rectangle section of sod for $200

Trava [24]

This isn't a question. What is the question?

3 0

3 years ago

At a high school, students can choose between three art electives, four history electives, and five computer electives. Each stu

V125BC [204]

Answer:

$\frac{^3C_1\times ^4C_1}{^{12}C_2}$

Step-by-step explanation:

Given,

Art electives = 3,

History electives = 4,

Computer electives = 5,

Total number of electives = 3 + 4 + 5 = 12,

Since, if a student chooses an art elective and a history elective,

So, the total combination of choosing an art elective and a history elective = $^3C_1\times ^4C_1$

Also, the total combination of choosing any 2 subjects out of 12 subjects = $^{12}C_2$

Hence, the probability that a student chooses an art elective and a history elective = $\frac{\text{Total combination of choosing an art elective and a history}}{\text{ Total combination of choosing any 2 subjects}}$

$=\frac{^3C_1\times ^4C_1}{^{12}C_2}$

Which is the required expression.

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3 years ago

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Solve the seperable equation:<br> dx/dt=3xt^2

andrew-mc [135]

$\displaystyle
\dfrac{dx}{dt}=3xt^2\\
\dfrac{dx}{x}=3t^2 \, dt\\
\int \dfrac{dx}{x}=\int 3t^2 \, dt\\
\ln x=3\dfrac{t^3}{3}+C\\
\ln x=t^3+C\\
x=e^{t^3+C}\\
x=e^{t^3}\cdot e^C\\
\boxed{x=Ce^{t^3}}$

5 0

2 years ago