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2 years ago

Which phrase best describes the translation from the group y=2x^2 to the graph of y= 2x^2 +5

Mathematics

1 answer:

dsp732 years ago

7 0

Answer:

The graph move up 5.

Step-by-step explanation:

When we add a constant to the end of a parabola, it shifts the graph up and down. Since the constant here is positive, the graph moves up by that number (5)

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Find the volume of the cylinder

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Answer:

$14130 {ft}^{3}$

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2 years ago

Which number line correctly shows 1.5+2.5?

Y_Kistochka [10]

Answer:

The first one!

Step-by-step explanation:

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2 years ago

Read 2 more answers

Pedro is doing math exercises from a software program he recently purchased. In the program, you have to get at least 70% of the

Nookie1986 [14]

The inequality would be
0.75q ≥ 0.70(q+1)

q is defined as the number of questions he answers after the first one. We are told he gets 75% of those correct; 75%=75/100=0.75. This gives us 0.75q.

Since he gains proficiency on the exercises, the total number he gets correct has to be at least 70%. This means the inequality would have the symbol greater than or equal to, as it cannot be less and have him gain proficiency.

He has already answered 1 question and answers q more; this gives us a total of q+1. Since he gains proficiency, the cutoff was 70%; 70%=70/100=0.70. This gives us the expression 0.70(q+1).

Our total inequality would then be 0.75q ≥ 0.70(q+1)

3 0

2 years ago

Read 2 more answers

Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1

aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

${3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}$

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

$3y^{2}.\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x}{y^{5}}$

3 0

2 years ago

A field is a rectangle with a perimeter of 1220 feet. The length is 100 feet more than the width. Find the width and length of t

rosijanka [135]

To start, let x represent the width and x+100 represent the length. Since the perimeter of a figure is the sum of all the measurements of the side which can be represented by (x+100)+(x+100)+x+x and since you know your perimeter is 1220, you can set the expression equal to 1220. This would look like this:
(x+100)+(x+100)+x+x=1220

Once you have done that, combine any like terms (combine terms with the same variables and raised to the same power together) which would simplify to this:
4x+200=1220

Now that you have your like terms simplified, subtract 200 from both sides to get 4x=1020 and finally, to solve for x, or find the width, divide both sides by 4 to get x=255.

Now that you have your width, now you must find your length as the question asks to find the dimensions of the rectangular field. To find the length, add 100 to the width, 255 since according to the information given, the length is 100 more than the width. When you add 100 to 255, you should get that your length is 355.

Now that you have your length and width, you can conclude that the dimensions of the field is 255 by 355 feet, which is your answer :)

3 0

2 years ago