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iragen [17]
3 years ago
13

All real numbers more than 4 units from 6 (in an absolute value inequality)

Mathematics
1 answer:
maks197457 [2]3 years ago
8 0
The correct answer for this question is "<span>|x-6| > 4."

</span>All real numbers more than 4 units from 6 (in an absolute value inequality).
We need to translate this sentence into a mathematical inequality.

"... units from 6 ..."
This means that the distance from 6 to a certain number.
x - 6

Since, we are talking about distance, place it inside an absolute symbol.
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Please help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
mash [69]

Answer:

4

Step-by-step explanation:

Multiply both sides by 7.

5s+8=28

Subtract 8 from both sides.

5s=20

Divide 5 from both sides,

s=4

3 0
3 years ago
Read 2 more answers
Pls help asap the best I can do is 50 points
lutik1710 [3]

Step-by-step explanation:

8.a.

a function is something that looks like

y = ax + b or similar.

it describes the way to calculate the result for whatever input value you can come up with.

so,

C = 12×d + 8

you could also say

C(d) = 12×d + 8

as, when you rent, you have to pay these $8 in any case, no matter how long you stay, and you need to pay $12 per day you stay.

funny thing is : you understood this perfectly well when doing b.

c.

the constant rate of change is always the factor of the input variable. here : 12

it is the ratio of (y change) / (x change)

or here (C change / d change).

it indicates that for any increase of d by 1 (staying one day longer) the charges C increase by 12.

9.a.

remember, what we just said about the rate of change ?

it is (result change) / (input change).

between 2010 and 2015

the input changed by +5 (from 2010 to 2015).

the result changed by +6.25 (from 37 to 43.25).

the rate of change is therefore

6.25/5 = 1.25 = 1 1/4 = 5/4

the function is therefore

population in millions = (years since 2010)×5/4 + 37

p = y×5/4 + 37

b.

now we want to find the number of years (y) so that the result of that equation is 50.

50 = y×5/4 + 37

13 = y×5/4

52 = 5y

y = 52/5 = 10.4

so, this will be during the 11th year after 2010 = 2021.

10.

be careful and concentrate.

you see that the sequence goes up 20, when n goes up 5 (and not just 1) ? there was your initial mistake.

a.

again, remember the change rate : it is result change / input change.

so, we have here 20/5 = 4.

so,

C(n) = 4×n + 50

b.

n = 12

C(12) = 4×12 + 50 = 98

c.

now we have C(n), and want to find for which n this is true.

C(n) = 78

78 = 4×n + 50

28 = 4×n

n = 7

after 7 visits the costs are 78.

there are the basic costs of 50 (for n = 0 or 0 visits) and then extra 4 for every visit.

d.

the x-axis is for the number of visits, the y-axis for the costs.

so, the line starts at the point (0, 50), and then passes through the points of the original table (5, 70), (10, 90), (15, 110). and if you want you can calculate the points for higher x (or number of visits).

4 0
3 years ago
A number a decreased by 7 is 2.
umka21 [38]

Answer:

9

Step-by-step explanation:

Let's first convert this to numbers. When a number is decreased by a certain amount, that is the same as saying that something is subtracted from it. Therefore:

a-7=2

Add 7 to both sides:

a-7+7=2+7

a=9

Hope this helps!

7 0
3 years ago
Read 2 more answers
((sinB)/1+cosB) + ((1+cosB)/sinB) = 2cscB prove that the equation is an identity
Contact [7]

Step-by-step explanation:

If you need any explanation, we can communicate normally

4 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
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