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Lynna [10]
3 years ago
6

I need help ASAP!!

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
4 0

Answer:

i think its 6

Step-by-step explanation:

i dont know for sure

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What kind of number is -5? Is it rational or something like that?
babymother [125]
The number -5 is anything but natural number.
It is an integer (Z) ,a rational number (Q) and a real number (R) .

4 0
3 years ago
Read 2 more answers
1-Solve: m - 9 = -13
timofeeve [1]
1: m=-4
2: b=-4
3: -9=k
4: C=-11.4
5:X<10
4 0
3 years ago
Can i have some help
Andrews [41]

Answer: g(x) = 3/2 x + 9

Step-by-step explanation:

h(x) = -2/3 x - 1

perpendicular lines always have the opposite sign, reciprocal slope

so, the slope of the perpendicular line would be: m = 3/2

y = mx + b

y = 3/2 x + b

plug in (-4, 3) to find b

3 = 3/2 (-4) + b

3 = -6 + b

b = 9

y = 3/2 x + 9

g(x) = 3/2 x + 9

5 0
2 years ago
Which ordered pair is the solution to the system of equations below? 4x+3y=31 2y=22-3x (A) (4,5) (B) (5,4) (C) (6,2) (D) (7,1)
grandymaker [24]

Plug in the points x and y

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al
Furkat [3]

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

3 0
2 years ago
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