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tatuchka [14]
3 years ago
12

Find the value of x. 7 Х 3 Z x= [?] V Give your answer in simplest form.

Mathematics
1 answer:
mixas84 [53]3 years ago
4 0

Answer:

21

Step-by-step explanation:

y^2=7*3\\y^2=21\\y^2=\sqrt{21}

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What polynomial must be added to 3x2 + 4x + 7 to obtain the sum of 0?
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do (3x2+4x+7)+y=0

y=-3x^2-4x-7

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A January 2007 Gallup Poll question asked, "In general, do you think things have gotten better or gotten worse in this country i
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Qualitative variable

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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
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