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scoundrel [369]
3 years ago
5

PLEASE HELPPP, its timed. NO BOTS please i want serious answers.

Mathematics
1 answer:
Rzqust [24]3 years ago
6 0
CDAB

4.2 (convert to frac on ur own)

12.5=x

CB= 14.5

AB= 8.82
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Logan opened a savings account 21 years ago with a deposit of $3,471.52. The account has an interest rate of 3.1% compounded mon
notka56 [123]
P = $3,471.52, the principal
r = 3.1% = 0.031, annual ratr
n = 12, monthly compounding
t = 21 years

Note that n*t = 252.

The value after 21 years is
A = 3471.52*(1 + 0.031/12)²⁵²
   = $6,650.91

The interest earned is
6650.91 - 3471.52 = 3179.39

Answer: $3,179.39
8 0
3 years ago
6x+3=10 hihihsoudhasoidjsaid
Vlada [557]

Step-by-step explanation:

6x+3=10

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7 0
3 years ago
which transformation of the function f(x) is the rigid motion: 2f(x)+3, f(x/2)+1, 2f(x/2), or f(x-1/2)+3
Katarina [22]
Answer is choice D. The other choices involve dilations, or scalings, which basically make the graph taller or flatter depending on what the scale factor is. In the case of f(x-1/2)+3, we shift f(x) to the right 1/2 a unit and up 3. 
4 0
3 years ago
Read 2 more answers
For this graph , mark the statements that are true
Rainbow [258]

ANSWER

A. The domain is the set of all real numbers.(True)

D. The range is the set of all real numbers. (True)

EXPLANATION

The given graph represent a polynomial function.

Polynomial functions are defined for all real values of x.

Also x is defined for all real values of y.

The first and the last options, A and D, are true.

Options B and C are false.

4 0
3 years ago
Prove De Morgan's law by showing that each side is a subset of the other side by considering x ∈ A⎯⎯⎯ A ¯ ∩ B⎯⎯⎯ B ¯ .
adelina 88 [10]

Solution :

We have to prove that $\overline{A \cup B} = \overline{A} \cap \overline{B}$   (De-Morgan's law)

Let  $x \in \bar{A} \cap \bar{B}, $ then $x \in \bar{A}$ and $x \in \bar{B} $

and so $x \notin \bar{A}$ and $x \notin \bar{B} $.

Thus, $x \notin A \cup B$ and so $x \in \overline{A \cup B}$

Hence, $\bar{A} \cap \bar{B} \subset \overline{A \cup B}$   .........(1)

Now we will show that $\overline{A \cup B} \subset \overline{A} \cap \overline{B}$

Let $x \in \overline{A \cup B}$ ⇒ $x \notin A \cup B$

Thus x is present neither in the set A nor in the set B, so by definition of the union of the sets, by definition of the complement.

$x \in \overline{A}$ and  $x \in \overline{B}$

Therefore, $x \in \overline{A} \cap \overline{B}$ and we have $\overline{A \cup B} \subset \overline{A} \cap \overline{B}$  .............(2)

From (1) and (2),

$\overline{A \cup B} = \overline{A} \cap \overline{B}$

Hence proved.

3 0
3 years ago
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