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Ilia_Sergeevich [38]
3 years ago
14

Is there enough information given to find the value of X. Explain your reasoning.

Mathematics
1 answer:
Anna007 [38]3 years ago
4 0
With what you've jus given me, no. is there an equation or something to go along with this question?

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Juan wants to buy a house for $162,800. The rate of down payment is 20% of the cost of the house. What is the down payment? i as
Paha777 [63]

Answer:

I think its gonna be 162.860

6 0
3 years ago
Solve t^4-2t^3+24t^2=0​
AlexFokin [52]

Answer:

t=0, 6, -4

Step-by-step explanation:

first, you can factor out t² of the equation

t²(t²-2t+24)=0, then factor the trinomial

t²(t-6)(t+4)=0, now set each factor equal to zero

t²=0, so t = 0

t-6=0, so t = 6

t+4=0, so t = -4

If you graph it, you can also see the answers.

4 0
4 years ago
Are these lines perpendicular, parallel, or neither based off their slopes?
tatyana61 [14]

Answer:

parallel

Step-by-step explanation:

//

5 0
2 years ago
Read 2 more answers
What is the radius of a sphere with a volume of 419 in^3, to the nearest tenth of an<br> inch?
vlabodo [156]

Answer:

R= 4.6 in

Step-by-step explanation:

Lol

5 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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