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Solnce55 [7]
3 years ago
13

What is l-8l? -16 -8 8 16

Mathematics
1 answer:
Makovka662 [10]3 years ago
6 0

Answer:

8

Step-by-step explanation:

Absolute Value  (| |) is the number of spaces away from 0. -8 is 8 spaces away from 0 and 8 is 8 spaces away from 0.

Hope this helps plz  mark brainliest if correct :D

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NEED HELP ASAP!!! Which questions are statistical?
uranmaximum [27]
1,2,4,6,7 is the answer
5 0
3 years ago
The 8th grade is selling tickets to the 8th grade formal.The cost per ticket is $15.The decorations and food cost $600,and each
MArishka [77]

Answer:

Step-by-step explanation:

Each ticket is $15.  The number of tickets is what we are trying to solve for.  The class spends a certain amount of money to prepare for the formal.  They hope that the money they make in ticket sales is MORE than what they spend.  The expression that represents the number of tickets at $15 each is 15x, where x is the number of tickets.  They hope that the sales are greater than what they spend, so what we have so far is

15x >

Greater than what, though?  What do they spend?  They spend 600 for the food, so

15x > 600...

but they also have to print a certain, unknown number of tickets at .50 each.  The expression that represents the printing of each ticket is .5x (we can drop the 0; it doesn't change the answer or make it wrong if we drop it off).  So the cost for this affair is the food + the printing.

15x > 600 + .5x

Solve this inequality for x.  Begin by subtracting .5 from both sides to get

14.5x > 600 so

x > 41.3

Because we are not selling (or printing) .3 of a ticket, it's safe to say (and also correct!) that they need to sell (and print) 41 tickets.  If they sell 41 tickets, the profit is found by

15(41) > 600 + .5(41)

615 > 600

This means that at 41 tickets, they make a profit.  At 40 tickets, the inequality looks like this:

15(40) > 600 + .5(40) and

600 > 620.  This is not true, so 40 tickets isn't enough.

8 0
3 years ago
Find the area of the region enclosed by the graphs of the functions
Vaselesa [24]

Answer:

\displaystyle A = \frac{8}{21}

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> </u>

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

  • Terms/Coefficients
  • Functions
  • Function Notation
  • Graphing
  • Solving systems of equations

<u>Calculus</u>

Area - Integrals

Integration Rule [Reverse Power Rule]:                                                                 \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Rule [Fundamental Theorem of Calculus 1]:                                      \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Addition/Subtraction]:                                                          \displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

Area of a Region Formula:                                                                                     \displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx

Step-by-step explanation:

*Note:

<em>Remember that for the Area of a Region, it is top function minus bottom function.</em>

<u />

<u>Step 1: Define</u>

f(x) = x²

g(x) = x⁶

Bounded (Partitioned) by x-axis

<u>Step 2: Identify Bounds of Integration</u>

<em>Find where the functions intersect (x-values) to determine the bounds of integration.</em>

Simply graph the functions to see where the functions intersect (See Graph Attachment).

Interval: [-1, 1]

Lower bound: -1

Upper Bound: 1

<u>Step 3: Find Area of Region</u>

<em>Integration</em>

  1. Substitute in variables [Area of a Region Formula]:                                     \displaystyle A = \int\limits^1_{-1} {[x^2 - x^6]} \, dx
  2. [Area] Rewrite [Integration Property - Subtraction]:                                     \displaystyle A = \int\limits^1_{-1} {x^2} \, dx - \int\limits^1_{-1} {x^6} \, dx
  3. [Area] Integrate [Integration Rule - Reverse Power Rule]:                           \displaystyle A = \frac{x^3}{3} \bigg| \limit^1_{-1} - \frac{x^7}{7} \bigg| \limit^1_{-1}
  4. [Area] Evaluate [Integration Rule - FTC 1]:                                                    \displaystyle A = \frac{2}{3} - \frac{2}{7}
  5. [Area] Subtract:                                                                                               \displaystyle A = \frac{8}{21}

Topic: AP Calculus AB/BC (Calculus I/II)  

Unit: Area Under the Curve - Area of a Region (Integration)  

Book: College Calculus 10e

6 0
3 years ago
Find the slope of the line passing through the points (-6,2), (0,-6).
Serhud [2]

C. -4/3

slope =  \frac{(y _{2} - y _{1}) }{(x _{2} - x _{1} )}  \\  =  \frac{( - 6 - 2)}{(0 -( -  6)}  \\  =  \frac{ - 8}{ 6}  \\  =   - \frac{4}{3}

3 0
3 years ago
Consider the rational expression (IMAGE ATTACHED)
Keith_Richards [23]

Answer:

  • 3x² is a term in the numerator
  • x + 1 is a common factor
  • The denominator has 3 terms

Step-by-step explanation:

You can identify terms and count them before you start factoring. Doing so will identify 3x² as a term in the numerator, and will show you there are 3 terms in the denominator.

When you factor the expression, you get ...

  \dfrac{3x^2-3}{3x^2+2x-1}=\dfrac{3(x^2-1)}{(3x-1)(x+1)}=\dfrac{3(x-1)(x+1)}{(3x-1)(x+1)}

This reveals a common factor of x+1.

So, the above three observations are true of this rational expression.

3 0
4 years ago
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