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Marina CMI [18]
3 years ago
10

8 into 6482 what is the placement value​

Mathematics
2 answers:
VARVARA [1.3K]3 years ago
4 0

Answer:

The tens place

Step-by-step explanation:

place values:

6000

400

80

2

elena-s [515]3 years ago
3 0

Answer:

It is 80.

Step-by-step explanation:

It's pretty simple-> Thousands, Hundreds, Tens, Ones.

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Factor 15z^2+17z-18 please show work
siniylev [52]

There is no common factor between 15, 17 and 18 because 17 is prime.  

15z² +17z - 18 =  

d = 17² - 4.15.-18  

d = 289 + 1080  

d = 1369  

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z" = (-17 - 37) : 30  

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<><

7 0
3 years ago
The amount of money in a bank account increased by 15.8% over the last year. If the amount of money at the beginning of the year
Lostsunrise [7]

Answer:

The expression represents the amount of money in the bank account after the increase is: x=1.158n.

Step-by-step explanation:

With the information provided, you can say that the amount of money in the bank after the increase would be equal to multiplying the amount at the beginning of the year for the result of adding up 1 plus the percentage of the increase, which would be:

x=n*(1+0.158)

x=n*1.158

x=1.158n, where:

x is the amount of money in the bank account after the increase

n is the amount of money at the beginning of the year

5 0
3 years ago
How can i differentiate this equation?
Dmitry_Shevchenko [17]

\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}


\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}


\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}

8 0
3 years ago
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