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3 years ago

Round this number to the nearest 1000 9372282

2 answers:

8 0

The answer would be 9372000 because when you round it to the nearest 1000 you would look at the 2 and see to the right if it is higher than 5 or lower then 4 and if it's not higher you would write zero after then two

kondor19780726 [428]3 years ago

4 0

9372 it stays the same

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How do round a decimal of 2.25 into a whole number ??

Mama L [17]

Answer:

Step-by-step explanation:

The whole number will be 2 because 2.25 is closer to 2 than 3 when we round to whole number

8 0

3 years ago

Plzzz help asapppppp what is the inverse of f if f(x)=^3 sqrt x-5

jasenka [17]

Hello from MrBillDoesMath!

Answer:

The first choice, y = x^3 + 5

Discussion:

Let

y = (x-5) ^ (1/3)

To find the inverse, sway x and y and then solve for y:

x = ( y - 5 )^(1/3) => cube both sides

x^3 = y - 5 => add 5 to both sides

x^3 + 5 = y

So the inverse is x^3 + 5, which is the first choice

Thank you,

MrB

4 0

3 years ago

Read 2 more answers

Please help with this problem:

Olegator [25]

He added 8 to both sides. He did the write thing.

6 0

3 years ago

Factorising quadratics 2nd page

mihalych1998 [28]

A^2-b^2=(a+b)(a-b)

1: x^2-4=(x+2)(x-2)
2: (x+8)(x-8)
3: (x+10)(x-10)
4: (x+14)(x-14)

7 0

3 years ago

4x2 + 4x - 1, evaluate and fully simplify each of the - For the function f(0) following f(s + h) f(x + h) - f(5) h 10

Andrew [12]

Given the function f(x);

$f(x)=-4x^2+4x-1$

Evaluating the function f(x+h);

$\begin{gathered} f(x+h)=-4(x+h)^2+4(x+h)-1 \\ f(x+h)=-4(x^2+2xh+h^2)^{}+4(x+h)-1 \\ f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1 \end{gathered}$

So;

$f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1$

Evaluating the second function;

$\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1-(-4x^2+4x-1)}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1+4x^2-4x+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2+4x^2-4h^2-8xh^{}+4x-4x+4h-1+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h} \end{gathered}$

simplifying further;

$\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h}=-4h-8x+4 \\ \frac{f(x+h)-f(x)}{h}=-4h-8x+4 \end{gathered}$

Therefore, we have;

$undefined$

7 0

1 year ago