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I am Lyosha [343]
3 years ago
7

5x(8-4)-4=

s (8 - 4) - 4 = " alt="5 \times (8 - 4) - 4 = " align="absmiddle" class="latex-formula">
Mathematics
2 answers:
MatroZZZ [7]3 years ago
4 0

The answer is 16! Hope this helps!


Svet_ta [14]3 years ago
3 0

5 x (8 - 4) - 4

= 5 x (4) - 4

= 20 - 4

= 16 <=== answer

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Simplify by combining like terms: 5x + 3x + 10x
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Step-by-step explanation:

Since all 3 are monomial terms with the same variable, we can add them together:

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Solve 5x^2 - 7x+2 -0 by completing the square. What are the solutions?
trapecia [35]

5<em>x</em>² - 7<em>x</em> + 2 = 0

5(<em>x</em>² - 7/5 <em>x</em>) + 2 = 0

5(<em>x</em>² - 7/5 <em>x</em> + 49/100 - 49/100) + 2 = 0

5(<em>x</em>² - 2 • 7/10 <em>x</em> + (7/10)²) - 49/20 + 2 = 0

5(<em>x</em> - 7/10)² - 9/20 = 0

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(<em>x</em> - 7/10)² = 9/100

<em>x</em> - 7/10 = ± √(9/100)

<em>x</em> - 7/10 = ± 3/10

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6 0
3 years ago
1.Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by
Genrish500 [490]

If x is a real number such that x3 + 4x = 0 then x is 0”.Let q: x is a real number such that x3 + 4x = 0 r: x is 0.i To show that statement p is true we assume that q is true and then show that r is true.Therefore let statement q be true.∴ x2 + 4x = 0 x x2 + 4 = 0⇒ x = 0 or x2+ 4 = 0However since x is real it is 0.Thus statement r is true.Therefore the given statement is true.ii To show statement p to be true by contradiction we assume that p is not true.Let x be a real number such that x3 + 4x = 0 and let x is not 0.Therefore x3 + 4x = 0 x x2+ 4 = 0 x = 0 or x2 + 4 = 0 x = 0 orx2 = – 4However x is real. Therefore x = 0 which is a contradiction since we have assumed that x is not 0.Thus the given statement p is true.iii To prove statement p to be true by contrapositive method we assume that r is false and prove that q must be false.Here r is false implies that it is required to consider the negation of statement r.This obtains the following statement.∼r: x is not 0.It can be seen that x2 + 4 will always be positive.x ≠ 0 implies that the product of any positive real number with x is not zero.Let us consider the product of x with x2 + 4.∴ x x2 + 4 ≠ 0⇒ x3 + 4x ≠ 0This shows that statement q is not true.Thus it has been proved that∼r ⇒∼qTherefore the given statement p is true.

8 0
2 years ago
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