All categories

3 years ago

PLZ NEED HELP ASAP WILL GIVE BRAINLIEST

Mathematics

1 answer:

german3 years ago

5 0

Answe quik tha 2nd one

Step-by-step explanation:

do it

You might be interested in

Check Register Balance $526.13 Bank Statement Balance $531.38 Outstanding Checks $62.45 and $18.75 Service Charges $75.95 What i

mrs_skeptik [129]

WOULD IT BE THE FIRST ONE ($526.13)

7 0

3 years ago

Which letter represents the y-intercept in the slope-intercept equation of a line: y = mx + b?

jeka94

Y = mx + b
m represents the slope
b represents the y intercept

8 0

3 years ago

How do I answer other people’s questions

iris [78.8K]

Click the answer button under there question

3 0

3 years ago

Read 2 more answers

Solve the inequality 8x > 48

Yakvenalex [24]

Step-by-step explanation:

$= 8x > 48$

Now , divide both side by the same number...

$= 8x \div 8 > 48 \div 8$

$= x > 6$

5 0

2 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

3 years ago