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stepan [7]
3 years ago
7

Faisal tries to solve

Mathematics
2 answers:
kramer3 years ago
8 0

Step-by-step explanation:

the steps he took are correct but the answer of x is somehow wrong because

(x+2)=0

x = -2

and not x = 2

Ronch [10]3 years ago
5 0

Answer:

He should have subtracted 2 from each side to find the solution for x+2 =0

Step-by-step explanation:

(x + 2)(x - 7) = 0

Using the zero product property

x+2 = 0   x-7=0

x+2-2=0-2    x-7+7=0+7

x = -2            x=7

He should have subtracted 2 from each side to find the solution for x+2 =0

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Is 1/7 greater than -5​
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no

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1/7 is positive, -5 is negative. Negative is less than positive

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Use the Pythagorean Theorem to find the length ​
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10

Step-by-step explanation:

64 +36=100

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Measure of the angle
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In total it has to be 180 since a straight line is 180 so i would be 69
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F(x) = x2 – 3x + 5 g(x) = V22 – 2x Calculate g (f (1)).​
erik [133]

Step-by-step explanation:

So first of all we plug in 1 into f(x) and the result of that into g(x).

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3 0
3 years ago
(−5)3x7(yz)4 / (3)2x2y8z2
g100num [7]

Answer:

\dfrac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}=\dfrac{-5x^5z^2}{2y^4}

Step-by-step explanation:

Given:

The expression to simplify is given as:

\frac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}

In order to simplify this, we have to use the law of indices.

1. (ab)^m=a^mb^m

So, (yz)^4=y^4z^4

Substitute this value in the above expression. This gives,

=\dfrac{(-5)3x^7y^4z^4}{(3)2x^2y^8z^2}\\\\\\=\dfrac{-15x^7y^4z^4}{6x^2y^8z^2}......(-5\times 3=15\ and\ 3\times 2=6)

Now, we use another law of indices.

2. \frac{a^m}{a^n}=a^{m-n}

So,  \frac{x^7}{x^2}=x^{7-2}=x^5,\frac{y^4}{y^8}=y^{4-8}=y^{-4}, \frac{z^4}{z^2}=z^{4-2}=z^2

Substitute these values in the above expression. This gives,

=\frac{-15}{6}\times x^5\times y^{-4}\times z^2\\\\=\frac{-5x^5y^{-4}z^2}{2}

Finally, we further simplify it using the law a^{-m}=\frac{1}{a^m}

So, y^{-4}=\frac{1}{y^4}

Therefore, the given expression is simplified as:

\dfrac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}=\dfrac{-5x^5z^2}{2y^4}

5 0
3 years ago
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