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3 years ago

Consider the sequence of steps to solve the equation: 3(x − 4) + 5x = 9x − 36

Mathematics

1 answer:

castortr0y [4]3 years ago

3 0

Answer:

The additive property of equality.

Step-by-step explanation:

36 is added to both sides of the equation.

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Select the correct answer.

musickatia [10]

Answer:

Step-by-step explanation:

3 - 7 = -4 and 3(3) + 2(-7) = -5, so D is the only choice that is correct for both inequalities.

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3 years ago

Six nickels is what percent of one dollar? What percent of $2.00 would it be? Please help.

Deffense [45]

$6\ nickels=6\times\ \$0.05=\$0.30\\\\\frac{0.03}{1}\cdot100\%=3\%\\\\6\ nickels\ is\ 3\%\ of\ \$1.\\\\\frac{0.03}{2}\cdot100\%=1.5\%\\\\6\ nickels\ is\ 1.5\%\ of\ \$2.$

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3 years ago

Translate this phrase into an algebraic expression

cupoosta [38]

Answer:

19x - 6

Step-by-step explanation:

Hope This Helps :)

Plz Mark Brainliest if Correct!!!

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3 years ago

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Blababa [14]

Hopefully these are correct lol, make sure to check it with a classmate. but i tried :)

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3 years ago

Consider the function V=g(x), where g(x) =x(6-2x)(8-2x), with x being the length of a cutout in cm and V being the volume of an

Andrej [43]

Answer:

The maximum volume of the open box is 24.26 cm³

Step-by-step explanation:

The volume of the box is given as $V=g(x)$ , where $g(x)=x(6-2x)(8-2x)$ and $0\le x\le3$ .

Expand the function to obtain:

$g(x)=4x^3-28x^2+48x$

Differentiate wrt x to obtain:

$g'(x)=12x^2-56x+48$

To find the point where the maximum value occurs, we solve

$g'(x)=0$

$\implies 12x^2-56x+48=0$

$\implies x=1.13,x=3.54$

Discard x=3.54 because it is not within the given domain.

Apply the second derivative test to confirm the maximum critical point.

$g''(x)=24x-56$ , $g''(1.13)=24(1.13)-56=-28.88\:$

This means the maximum volume occurs at $x=1.13$ .

Substitute $x=1.13$ into $g(x)=x(6-2x)(8-2x)$ to get the maximum volume.

$g(1.13)=1.13(6-2\times1.13)(8-2\times1.13)=24.26$

The maximum volume of the open box is 24.26 cm³

See attachment for graph.

6 0

3 years ago