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Makovka662 [10]
3 years ago
9

What is 218,742 rounded to the nearest hundred thousand?

Mathematics
2 answers:
Anastasy [175]3 years ago
5 0

Answer:219,000

Step-by-step explanation: 7 is higher than 5 so you round it positively :)

xxMikexx [17]3 years ago
4 0

Answer:

200,000

Step-by-step explanation:

its easy and simple lol

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13,000 divided by 1,000
ivann1987 [24]

Answer:

Hey! The answer would be 13 with no remainder. Hope this helped :)

3 0
3 years ago
alex spent 3/7 of his money he gave 1 /4 of the remainder to his sistet he haf $120 left how much money did he in the beginning
Sonbull [250]
Let his total money be x.
He spent (3/7)*x of his money.
He gave 1/4(x-3/7x) to his sister. 
which is : (1/4)*(4/7)*x = (1/7)*x 
Now,
How much money he have in the beginning = 
(3/7)*x+(1/7)*x+120=x
(4/7)*x+120=x
120=x-(4/7)*x
120=(3/7)*x
x=120*7/3 [cross multiplication]
x=280.

So, Alex had $280 in the beginning.

6 0
3 years ago
-3.5 + x7= -5.2<br><br><br> pls answer
noname [10]
X=-17/70

your welcome
8 0
3 years ago
Find the negation of the proposition
murzikaleks [220]

In accordance with <em>propositional</em> logic, <em>quantifier</em> theory and definitions of <em>simple</em> and <em>composite</em> propositions, the negation of a implication has the following equivalence:

\neg (\exists \,x\, (P(x) \implies  Q(x))) \iff \forall \,x (\neg \,Q(x) \,\land \,P(x)) (Correct choice: iii)

<h3>How to find the equivalent form of a proposition</h3>

Herein we have a <em>composite</em> proposition, that is, the union of <em>monary</em> and <em>binary</em> operators and <em>simple</em> propositions. According to <em>propositional</em> logic and <em>quantifier</em> theory, the negation of an implication is equivalent to:

\neg (\exists \,x\, (P(x) \implies  Q(x))) \iff \forall \,x (\neg \,Q(x) \,\land \,P(x))

To learn more on propositions: brainly.com/question/14789062

#SPJ1

8 0
2 years ago
Solve 5x^2-2=-12 by taking the square root
k0ka [10]

Answer:

x = ±i√2

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  

Equality Properties

Multiplication Property of Equality

Division Property of Equality

Addition Property of Equality

Subtraction Property of Equality<u> </u>

<u>Algebra II</u>

Imaginary root <em>i</em>

  • <em>i</em> = √-1

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

5x² - 2 = -12

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. [Addition Property of Equality] Add 2 on both sides:                                    5x² = -10
  2. [Division Property of Equality] Divide 5 on both sides:                                 x² = -2
  3. [Equality Property] Square root both sides:                                                    x = ±√-2
  4. Rewrite:                                                                                                             x = ±√-1 · √2
  5. Simplify:                                                                                                             x = ±i√2
7 0
3 years ago
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