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2 years ago

10

Helppppp pleasee....

1 answer:

irina1246 [14]2 years ago

6 0

Independent variable: the variable that affects the dependent variable and creates a change
Dependent variable: depends on the independent variable

1. Independent variable: John’s age
Dependent variable: height

2. Independent: gallons of milk
Dependent: cups of milk

3. Independent: population
Dependent: number of representatives

4. Independent: Number of seats
Dependent: amount of tickets

5. Independent: time
Dependent: plane’s altitude

6. Independent: more music download
Dependent: more money spent

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How to work the problem 2u²=3u+2

vlabodo [156]

U= -1/2, 2

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2 years ago

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Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in t

Tresset [83]

Answer: The volume of largest rectangular box is 4.5 units.

Step-by-step explanation:

Since we have given that

Volume = $xyz$

with subject to $x+2y+3z=9$

So, let $z=\dfrac{9-x-2y}{3}$

So, Volume becomes,

$V=xyz\\\\V=xy(\dfrac{9-x-2y}{3})\\\\V=\dfrac{9xy-x^2y-2xy^2}{3}$

Partially derivative wrt x and y we get that

$9-2x-2y=0\implies 2x+2y=9\\\\and\\\\9-x-4y=0\implies x+4y=9$

By solving these two equations, we get that

$x=3,y=\dfrac{3}{2}$

So, $z=\dfrac{9-x-2y}{3}=\dfrac{9-3-3}{3}=\dfrac{3}{3}=1$

So, Volume of largest rectangular box would be

$xyz=3\times \dfrac{3}{2}\times 1=\dfrac{9}{2}=4.5$

Hence, the volume of largest rectangular box is 4.5 units.

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2 years ago

Is 5.692 a terminating or a repeating decimal

Ostrovityanka [42]

5.692 is a terminating decimal because the decimal stopped at the digit of 2.

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Amanda uses the equation y = kx to determine how much money (y) she earns after working any number of hours (x). Amanda earned $

stiks02 [169]

The answer would be A) 8.75

To solve you substitute the numbers into the equation y=kx
350=40k
then you would divide and get 8.75

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VikaD [51]

A prime number is a natural number greater than 1 that can't be the factory of multiplying two smaller natural numbers.

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