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gizmo_the_mogwai [7]
3 years ago
9

What is the product? Assume x20 ( 13x + 15 )( V15x+2/30 O 3x 15+3./165x + 10-16 O 3x V5+6 /10x+5-3x+10.56 0 3x √5 + 10/6 6./3x +

10 76​
Mathematics
1 answer:
miv72 [106K]3 years ago
6 0

Answer:

doesn't make sense

Step-by-step explanation:

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2(5 - x)>8 solve for x​
nikitadnepr [17]

Answer:

x < 1

Step-by-step explanation:

To solve for x, we need to get x by itself

2(5 - x)>8

Distribute the 2

2*5 -2*x >8

10- 2x >8

Subtract 10 from both sides

10-10 -2x >8 -10

-2x > -2

Divide both sides by -2

-2x/-2> -2/-2

Since we divided by a negative number, we must flip the inequality sign

x < 1

5 0
3 years ago
Read 2 more answers
A number line graph is shown:
ipn [44]

Answer:

-6 > - 4

Step-by-step explanation:

3 0
3 years ago
In 12 hours, the temperature fell steadily from 17degrees to -7degrees f. what was the average change in temperature per hour?
Zarrin [17]
If the temperature changed from 17 degree F to -7 , the change was
17+7=24 degrees F
the average change was 24/12=2 , 2 degrees per hour 
<span />
7 0
3 years ago
PLEASE HELP!<br> Solve the open sentence. <br><br><br> 8 &lt; 2b or 2b + 15 &lt; 5
SIZIF [17.4K]
The answer: b><span><span><span>4<span> or </span></span>b</span><<span>−<span>5</span></span></span>
6 0
3 years ago
Let Ebe the set of all even positive integers in the universe Zof integers, and XE : Z R be the characteristic function of E.
AnnZ [28]

Answer:

\mathbf{X_E (2) =  1}

\mathbf{X_E (-2) = 0 }  

\mathbf{\{ x \in Z: X_E(x) = 1\}  = E}

Step-by-step explanation:

Let E be the set of all even positive integers in the universe Z of integers,

i.e

E = {2,4,6,8,10 ....∞}

X_E : Z \to R be the characteristic function of E.

∴

X_E(x) = \left \{ {{1 \ if  \ x \ \  is \ an \ element \ of \ E} \atop {0 \ if  \ x \ \  is \ not \ an  \ element \ of \ E}} \right.

For XE(2)

\mathbf{X_E (2) =  1}  since x is an element of E (i.e the set of all even numbers)

For XE(-2)

\mathbf{X_E (-2) = 0 }   since  - 2 is less than 0 , and -2 is not an element of E

For { x ∈ Z: XE(x) = 1}

This can be read as:

x which is and element of Z such that X is also an element of x which is equal to 1.

∴

\{ x \in Z: X_E(x) = 1\} = \{ x \in Z | x \in E\} \\ \\  \mathbf{\{ x \in Z: X_E(x) = 1\}  = E}

E = {2,4,6,8,10 ....∞}

5 0
3 years ago
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