All categories

2 years ago

What’s the standard form for 1,000,000+80,000+30+3

Mathematics

2 answers:

PilotLPTM [1.2K]2 years ago

6 0

<h3>Answer: 1,080,033</h3>

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

Explanation:

1,000,000 = 1 million

80,000 = 80 thousand

Adding those two would get us

1,000,000+80,000 = 1,080,000

Then we change those last two zeros at the end of "1,080,000" to 33 instead to get the final answer 1,080,033. This is done because we're adding 30+3 = 33 at the end.

----------------------

Extra info:

The number 1,080,033 is slightly over 1.08 million
1,080,033 = 1 million, 80 thousand, 33
1,080,033 = one million, eighty thousand, thirty-three
The use of commas between the words is optional. It's to give a sense of the pacing or pauses when reading out the number.

Alenkasestr [34]2 years ago

5 0

Answer:

1080033

Step-by-step explanation:

You might be interested in

Create a quadratic polynomial function f(x) and a linear binomial in the form (x − a).

Tamiku [17]

Our quadratic polynomial function is:

f(x) = 4x^2 + 2x + 5

our linear binomial form is:
x - a = -2
hence:
a = x + 2

and so we have:
f(x) = 4x^2 + 2x + 5
f(a) = 4(x + 2)^2 + 2<span>(x + 2)</span> + 5
= 4(x^2 + 4x + 4) + 2x + 4 + 5
f(a) = 4x^2 + 18x + 25

3 0

3 years ago

How to show the tens fact you used. write the difference. <br> 14-6=____<br> 10-___=____

Dominik [7]

This is what I believe you are looking for. 14-6=8
10-2=8

8 0

3 years ago

9. The Science Club needs to rent a bus for a field trip. Main Street Buses charges a $40 rental fee, plus $2 per mile. County B

Basile [38]

Answer:

20 miles $80

Step-by-step explanation:

20 miles

P%5Bc%5D%2820%29=20%2B3%2820%29=20%2B60=80

$80

8 0

3 years ago

Read 2 more answers

Which method will not work?

katen-ka-za [31]

D because if you multiply y 8 times it will no longer be doubling the area

6 0

3 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago