he just remember that number had four digits and a tree in the 10 Place write a number that he could be thinking of

Mathematics

1 answer:

viva [34]4 years ago

8 0

This what you gotta do 1+1=2 you get it

You might be interested in

What is a variable in mathematics?

Dmitrij [34]

Answer:

variable, In algebra, a symbol (usually a letter) standing in for an unknown numerical value in an equation. Commonly used variables include x and y (real-number unknowns), z (complex-number unknowns), t (time), r (radius), and s (arc length).

7 0

2 years ago

Read 2 more answers

What’s the answer for 12?

Lisa [10]

Answer:

$- 3 \sqrt{4 \times 4 \times 3} \\ - 3 \times 2 \times 2 \sqrt{3 } \\ - 12 \sqrt{3}$

6 0

3 years ago

Please show proof of how u got it ;-; :(

yanalaym [24]

First, divide the the polygon into two quadrilateral.

Then find the area of both quadrilaterals

The area of the small rectangle = 4 x 3 = 12 square feet

The area of the bigger rectangle = (9 - 3) x 14 = 6 x 14 = 84 square feet

Add both areas:

12 + 84 = 96 square feet

8 0

4 years ago

Please help ASAP

scZoUnD [109]

600x1.07^11 = 1262
11 Months

8 0

3 years ago

Evaluate the integral by changing to polar coordinatesye^x dA, where R is in the first quadrant enclosed by the circle x^2+y^2=2

34kurt

$\displaystyle\iint_Rye^x\,\mathrm dA=\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=5}r^2\sin\theta e^{r\cos\theta}\,\mathrm dr\,\mathrm d\theta$

which follows from the usual change of coordinates via

$\begin{cases}\mathbf x(r,\theta)=r\cos\theta\\\mathbf y(r,\theta)=r\sin\theta\end{cases}$

and Jacobian determinant

$|\det J|=\left|\begin{vmatrix}\mathbf x_r&\mathbf x_\theta\\\mathbf y_r&\mathbf y_\theta\end{vmatrix}\right|=|r|$

Swap the order, so that the integral is

$\displaystyle\int_{r=0}^{r=5}\int_{\theta=0}^{\theta=\pi/2}r^2\sin\theta e^{r\cos\theta}\,\mathrm d\theta\,\mathrm dr$

and now let $\sigma=r\cos\theta$ , so that $\mathrm d\sigma=-r\sin\theta$ . Now, you have

$\displaystyle\int_{r=0}^{r=5}\int_{\sigma=0}^{\sigma=r}re^\sigma\,\mathrm d\sigma\,\mathrm dr=\int_{r=0}^{r=5}r(e^r-1)\,\mathrm dr=4e^5-\dfrac{23}2$

7 0

4 years ago