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3 years ago

Solve for y: 11-1/2y=3+6x

2 answers:

7 0

$11-\dfrac{1}{2}y=3+6x\ \ \ \ \ |subtract\ 11\ from\ both\ sides\\\\-\dfrac{1}{2}y=-8+6x\ \ \ \ \ |change\ signs\\\\\dfrac{1}{2}y=8-6x\ \ \ \ \ |multiply\ both\ sides\ by\ 2\\\\y=4-3x$

jonny [76]3 years ago

6 0

Answer:

The value of the provided equation for y is $y=-12x+16$ .

Step-by-step explanation:

Consider the provided equation.

$11-\frac{1}{2}y=3+6x$

We need to solve the above equation for y.

Subtract 11 from both the sides.

$11-11-\frac{1}{2}y=3-11+6x$

$-\frac{1}{2}y=6x-8$

Multiply both the sides by -2

$-\frac{1}{2}y\times -2=-2(6x-8)$

$y=-12x+16$

Hence, the value of the provided equation for y is $y=-12x+16$ .

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I need some help with this calculus 1 implicit diffrention problem?

katen-ka-za [31]

Answer:

y" = -24 / y³

Step-by-step explanation:

6x² + y² = 4

Take the derivative of both sides with respect to x.

12x + 2y y' = 0

Again, take the derivative of both sides with respect to x.

12 + 2y y" + y' (2y') = 0

12 + 2y y" + 2(y')² = 0

Solve for y' in the first equation.

2y y' = -12x

y' = -6x/y

Substitute and solve for y":

12 + 2y y" + 2(-6x/y)² = 0

12 + 2y y" + 2(36x²/y²) = 0

12 + 2y y" + 72x²/y² = 0

6y² + y³ y" + 36x² = 0

y³ y" = -36x² − 6y²

y" = (-36x² − 6y²) / y³

Solve for y² in the original equation and substitute:

y² = 4 − 6x²

y" = (-36x² − 6(4 − 6x²)) / y³

y" = (-36x² − 24 + 36x²) / y³

y" = -24 / y³

7 0

3 years ago

Solve the percent problem 76 gallons is 126% of ___ gallons. Round to the nearest hundredth, if necessary.

Burka [1]

Answer:

60.32

Step-by-step explanation:

126%=1.26

60.32*1.26=76.0032

76.0032≈76

7 0

3 years ago

What is the solution to the equation sqrt n= sqrt 3n-3 +1?

inna [77]

The answer would be n = 1

3 0

4 years ago

I need help with this I got -10 is it right??

Dahasolnce [82]

Answer:

-14

Step-by-step explanation:

difference between -20 and -5 = 15

15 x 2/5 = 30/5 = 6

-20 + 6 = -14

7 0

3 years ago

A chemical company makes two brands of antifreeze. The first brand is 20% pure antifreeze, and the second brand is 70% pure anti

andre [41]

Answer:

First brand of antifreeze: 21 gallons

Second brand of antifreeze: 9 gallons

Step-by-step explanation:

Let's call A the amount of first brand of antifreeze. 20% pure antifreeze

Let's call B the amount of second brand of antifreeze. 70% pure antifreeze

The resulting mixture should have 35% pure antifreeze, and 30 gallons.

Then we know that the total amount of mixture will be:

$A + B = 30$

Then the total amount of pure antifreeze in the mixture will be:

$0.2A + 0.7B = 0.35 * 30$

$0.2A + 0.7B = 10.5$

Then we have two equations and two unknowns so we solve the system of equations. Multiply the first equation by -0.7 and add it to the second equation:

$-0.7A -0.7B = -0.7*30$