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Paha777 [63]
2 years ago
7

These tables represent a quadratic function with a vertex at (0, 3). What is the

Mathematics
1 answer:
astra-53 [7]2 years ago
7 0
The answer is d . -97
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I need help with this question, I can work out the ratio but I am just confused with the question. So do help if you can please.
nirvana33 [79]

Answer:

76%

Step-by-step explanation:

So basically 528 represents a full circle and the amount in the circle is 2:5 to that in the stalls.

So you divide 528 by 2 and then multiply it by 5 to get a full stall of 1320

you then add 1320 and 528 to get a full theater of 1848

to get 2/3 of seats in the stall you divide 1320 by 3 and then multiply it by 2 to get 880.

For 2/3 of the stall and a full circle you add 880 and 528 to get 1408.

You divide the amount of people in the theater on Friday by the total amount of people which the theater can hold so 1408 divided by 1848 and you get the percentage of 76%

8 0
3 years ago
PLEASE HELP ME BEST ANSWER GETS BRAINLIEST NO SCAMS PLEASE!
fomenos

Answer:

False

Step-by-step explanation:

People reading in a library are most likely going to like to read books

8 0
3 years ago
Read 2 more answers
Find the equation of a line perpendicular to -x=4-y that passes through the point (-3,8)
Wewaii [24]

9514 1404 393

Answer:

  y = -x +5

Step-by-step explanation:

Solving the given equation for y, we have ...

  y = x +4

The slope of this line is the coefficient of x: 1. The slope of the perpendicular line will be the opposite reciprocal of this: -1/1 = -1. The y-intercept of the perpendicular line can be found from ...

  b = y -mx = 8 -(-1)(-3) = 5

The perpendicular line has equation ...

  y = -x +5

8 0
2 years ago
If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form
algol13

Answer:

\frac{d^2y}{dx^2} = \frac{-4}{3}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Factoring

<u>Calculus</u>

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Product Rule: \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Chain Rule: \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Quotient Rule: \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

<u>Step 2: Differentiate Pt. 1</u>

<em>Find 1st Derivative</em>

  1. Implicit Differentiation [Basic Power Rule]:                                                  -y'-6x^2=2yy'
  2. [Algebra] Isolate <em>y'</em> terms:                                                                              -6x^2=2yy'+y'
  3. [Algebra] Factor <em>y'</em>:                                                                                       -6x^2=y'(2y+1)
  4. [Algebra] Isolate <em>y'</em>:                                                                                         \frac{-6x^2}{(2y+1)}=y'
  5. [Algebra] Rewrite:                                                                                           y' = \frac{-6x^2}{(2y+1)}

<u>Step 3: Differentiate Pt. 2</u>

<em>Find 2nd Derivative</em>

  1. Differentiate [Quotient Rule/Basic Power Rule]:                                          y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}
  2. [Derivative] Simplify:                                                                                       y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}
  3. [Derivative] Back-Substitute <em>y'</em>:                                                                     y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}
  4. [Derivative] Simplify:                                                                                      y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}

<u>Step 4: Find Slope at Given Point</u>

  1. [Algebra] Substitute in <em>x</em> and <em>y</em>:                                                                     y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}
  2. [Pre-Algebra] Exponents:                                                                                      y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}
  3. [Pre-Algebra] Multiply:                                                                                   y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}
  4. [Pre-Algebra] Add:                                                                                         y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}
  5. [Pre-Algebra] Exponents:                                                                               y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}
  6. [Pre-Algebra] Divide:                                                                                      y''(-1,-2) = \frac{-36+24 }{9}
  7. [Pre-Algebra] Add:                                                                                          y''(-1,-2) = \frac{-12}{9}
  8. [Pre-Algebra] Simplify:                                                                                    y''(-1,-2) = \frac{-4}{3}
6 0
3 years ago
Which construction is illustrated above?
Leni [432]

Answer:

The construction is a parallel line to a given line  from a point not on the line.

Step-by-step explanation:

For a given figure,

To construct a parallel line

Step 1 : Draw transversal line through any point not on line which is given.

Step 2: Using construction copy an angle, construct same angle formed by transversal line and given line.

Step 3: This is the last step in which when the copy of an angle completed draw a line parallel to the line.

6 0
3 years ago
Read 2 more answers
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