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Alekssandra [29.7K]
3 years ago
5

Help !!!!!!!!!!! me part 2 question read 1 then read 2 I andswer one need help with 2

Mathematics
1 answer:
sergiy2304 [10]3 years ago
7 0

Answer:

The volume of a cylinder is given by πr²h where, r is the radius of the cylinder and h is the height of the cylinder. Also r=d/2 , where d is the diameter of the cylinder. Therefore the volume becomes one-fourth of the initial volume. ... The new volume is 730 mL.

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Could someone explain how to do this, I have no idea how...
alexira [117]

Answer:

2a²

Step-by-step explanation:

Pair 'like' terms with 'like' terms, ie numbers go with numbers, and 'a's go with 'a's.

Lets deal with the top of the fraction first:

4ax3a³

Rearrange it so you have numbers beside numbers and 'a's beside 'a's:

(4x3)x(axa³)

12x(a⁴) <em>(because nᵃxnᵇ=nᵃ⁺ᵇ)</em>

12a⁴

Now, instead of (4ax3a³)/6a², we have 12a⁴/6a²

First divide the numbers: 12/6 =2

Now divide the 'a' parts: a⁴/a²=a² <em>(because nᵃ/nᵇ=nᵃ⁻ᵇ)</em>

Now we have 2a²

7 0
3 years ago
Remember to show work and explain. Use the math font.
MrMuchimi

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

6 0
3 years ago
Which of ordered pairs is not a function ?
Free_Kalibri [48]
Answer:4)\\\\because\\(2;\ 3)\to x=2\ and\ y=3\\(2;\ 4)\to x=2\ and\ y=4\\\\for\ this\ same\ "x"\ y=3\ and\ y=4\\\\and\\\\(4;\ 5)\to x=4\ and\ y=5\\(4;\ 6)\to x=4\ and\ y=6\\\\for\ this\ same\ "x"\ y=5\ and\ y=6
4 0
3 years ago
Find variable values for two variables in an equation
never [62]

Simplifying

2x + 3y = 12

Solving

2x + 3y = 12

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-3y' to each side of the equation.

2x + 3y + -3y = 12 + -3y

Combine like terms: 3y + -3y = 0

2x + 0 = 12 + -3y

2x = 12 + -3y

Divide each side by '2'.

x = 6 + -1.5y

Simplifying

x = 6 + -1.5y

5 0
3 years ago
Use the functions and the scenario to solve the problem.
Naily [24]

Answer:

Leanne is correct. It is important that g(x) is subtracted from f(x).

Shiloh is incorrect. She subtracted f(x) from g(x), and subtraction isn't commutative.

*Just took the test, got the answer correct!! Hope this helps!

7 0
2 years ago
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