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den301095 [7]
3 years ago
5

What number makes the equation true ÷3=6

Mathematics
1 answer:
Akimi4 [234]3 years ago
7 0

Answer:2

Step-by-step explanation:

you just  do 6 divided by 3 and get 2

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Six less than 2 times “y” is 34 equation form
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2y-6=34 is the needed equation
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James determined that these two expressions were equivalent expressions using the values of x = 4 and x = 6. Which statements ar
GaryK [48]

Answer:

The expressions 7x + 4 and 3x + 5 + 4x = 1 are equivalent for every value of x.

Step-by-step explanation:

Expressions - 7x + 4 and 3x + 5 + 4x = 1.

1. When x = 2, both expressions have a value of 18. Let's find out:

1st expression : 7(2) + 4 = 14+4 = 18

2nd expression : 3(2) + 5 + 4(2) = 1

or 6 + 5 + 8 - 1 = 18

Values of both the expressions is 18, hence it is correct statement.

2.The expressions are only equivalent for x = 4 and x = 6.

This is incorrect statement as we just calculated above that the expressions are equivalent for x = 2. Hence, it is incorrect.

3. The expressions have equivalent values for any value of x.

Say, x = 0, then,

7x + 4 = 7 (0) + 4 = 4 and,

3x + 5 + 4x - 1 =  3(0) + 5 + 4(0) - 1  = 5-1 = 4

The statement holds.

Let's try again for x = 12,

7x + 4 = 7(12) + 4 = 88 and,

3x + 5 + 4x - 1 =  3(12) + 5 + 4(12) - 1  = 36 + 5 + 48 -1 =  88

Let's try again for x = 13,

7x + 4 = 7(13) + 4 = 95 and,

3x + 5 + 4x - 1 =  3(13) + 5 + 4(13) - 1  = 39 + 5 + 52 -1 =  95

Clearly, it holds for every value of x, whether it is odd or even. Hence, it is correct statement.

_______________________________________________________

Trick: 7x + 4 = 0....(1) and,

3x + 5 + 4x = 1

Rearranging the terms of above expression, we get,

(3x + 4x) + (5 - 1) =0

or 7x + 4 = 0...(2)

clearly both (1) & (2) are equivalent.

_____________________________________________________

4 0
3 years ago
Simplify. square root (108 x^5 y^6)
sveta [45]
\bf \sqrt{108x^5y^6}\qquad 
\begin{cases}
108=2\cdot 2\cdot 3\cdot 3\cdot 3\\
\qquad 2^2\cdot 3^2\cdot 3\\
\qquad (2\cdot 3)^2\cdot 3\\
\qquad 6^2\cdot 3\\
x^5=x^{4+1}\\
\qquad x^4\cdot  x^1\\
\qquad x^{2\cdot 2}\cdot x\\
\qquad (x^2)^2\cdot x\\
y^6=y^{3\cdot 2}\\
\qquad (y^3)^2
\end{cases}\implies \sqrt{6^2\cdot 3\cdot (x^2)^2\cdot x\cdot (y^3)^2}
\\\\\\
6x^2y^3\sqrt{3x}
3 0
3 years ago
Find the area of Pentagon ABCDE in which BL perpendicular to AC, CM perpendicular to AD, EN perpendicular to AD such that AC = 8
deff fn [24]

Answer:

Area ABCDE = 82 cm²

Step-by-step explanation:

ABCDE = ΔEAD + ΔDAC + ΔCAB

ABCDE = (12×5)/2 + (12×6)/2 + (8×4)/2 = 30+36+16 = 82

6 0
3 years ago
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