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2 years ago

For which equation is x=0 NOT a solution?

1 answer:

4 0

B, is the only solution where x=0 isn’t the answer because 2.5x+3=x and if u subtract the 2.5x you get 3= -1.5x and if you divide 3 by -1.5 it won’t be x=0

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Find the greatest solution for x+y when x^2+y^2 = 7, x^3+y^3=10

damaskus [11]

Answer:

Step-by-step explanation:

set

$f(x,y)=x+y\\$

constrain:

$g(x,y)=x^2+y^2 = 7\\h(x,y)=x^3+y^3=10$

Partial derivatives:

$f_{x}=1\\f_{y} =1 \\g_{x}=2x \\g_{y}=2y\\h_{x}=3x^2 \\h_{y}=3y^2$

Lagrange multiplier:

$grad(f)=a*grad(g)+b*grad(h)\\$

$\left[\begin{array}{ccc}1\\1\end{array}\right]=a\left[\begin{array}{ccc}2x\\2y\end{array}\right]+b\left[\begin{array}{ccc}3x^2\\3y^2\end{array}\right]$

4 equations:

$1=2ax+3bx^2\\1=2ay+3by^2\\x^2+y^2=7\\x^3+y^3=10$

By solving:

$a=4/9\\b=-2/27\\x+y=4$

Second mathod:

Solve for x^2+y^2 = 7, x^3+y^3=10 first:

$x=\frac{1}{2} -\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} +\frac{\sqrt{13}}{2} \\x=\frac{1}{2} +\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} -\frac{\sqrt{13}}{2} \\x+y=-5\ or\ 1 \or\ 4$

The maximum is 4

6 0

2 years ago

Please help. I don't need a big explanation just a answer

stiv31 [10]

A answer of the explanation for the explanation is the answer

6 0

2 years ago

4(a + 2) = 14 – 2(3 – 2a) –2 –1 no solution all real numbers

Ray Of Light [21]

All real numbers.

$\boxed{TRUE}$

Step-by-step explanation:

Distributive property: a(b+c)=ab+ac

Expand: 4(a+2)=4a+8

Expand again: 14-2(3-2a)=4a+8

4a+8=4a+8

You subtract by 8 from both sides of an equation.

4a+8-8=4a+8-8

Simplify.

4a=4a

Then, subtract by 4a from both sides of an equation.

4a-4a=4a-4a

Finally, simplify.

4a-4a=0

0=0

True

All real numbers is the final answer.

Hope this helps you!

Have a nice day! :)

4 0

2 years ago

(100 pts.)

ICE Princess25 [194]

Answer:

1260 is the correct answer

remember multiply them :)

Step-by-step explanation:

7 0

2 years ago

I am giving branliest

Oduvanchick [21]

Answer:

eisdmejdjdj

Step-by-step explanation:

kdjeirkffn

4 0

2 years ago