1. I am guessing the triangles split the bottom in half so you will find the area of the triangles and the rectangle and add them together
Triangle=3×4 /2 12/2=6
Triangle 3×4 /2 6
Rectangle: 8×1 8
6+6+8=
12+8=20m^2
2. Triangle area+rectangle area
Rectangle: 22×24 528
Triangle: 8×24 /2 192/2 96
528+96= 624
1) Multiply everything by 8. It clears the fractions.
It gives you 6n + 128 =16 - n
You add the -n to the other side of the equation.
It gives you 7n + 128= 16
You then subtract 128 from 16.
It gives you 7n= -112
You divide everything by 7
It gives you n=-17
If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301—————
Solve the initial value problem:
dy——— = 2xy², y = 2, when x = – 1. dxSeparate the variables in the equation above:
![\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\ \mathsf{y^{-2}\,dy=2x\,dx}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7By%5E2%7D%3D2x%5C%2Cdx%7D%5C%5C%5C%5C%0A%5Cmathsf%7By%5E%7B-2%7D%5C%2Cdy%3D2x%5C%2Cdx%7D)
Integrate both sides:
![\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\ \mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\ \mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\ \mathsf{-\,\dfrac{1}{y}=x^2+C_1}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21y%5E%7B-2%7D%5C%2Cdy%3D%5Cint%5C%212x%5C%2Cdx%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7By%5E%7B-2%2B1%7D%7D%7B-2%2B1%7D%3D2%5Ccdot%20%5Cdfrac%7Bx%5E%7B1%2B1%7D%7D%7B1%2B1%7D%2BC_1%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7By%5E%7B-1%7D%7D%7B-1%7D%3D%5Cdiagup%5Chspace%7B-7%7D2%5Ccdot%20%5Cdfrac%7Bx%5E2%7D%7B%5Cdiagup%5Chspace%7B-7%7D2%7D%2BC_1%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B-%5C%2C%5Cdfrac%7B1%7D%7By%7D%3Dx%5E2%2BC_1%7D)
![\mathsf{\dfrac{1}{y}=-(x^2+C_1)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7B1%7D%7By%7D%3D-%28x%5E2%2BC_1%29%7D)
Take the reciprocal of both sides, and then you have
![\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}](https://tex.z-dn.net/?f=%5Cmathsf%7By%3D-%5C%2C%5Cdfrac%7B1%7D%7Bx%5E2%2BC_1%7D%5Cqquad%5Cqquad%20where~C_1~is~a~constant%5Cqquad%20%28i%29%7D)
In order to find the value of
C₁ , just plug in the equation above those known values for
x and
y, then solve it for
C₁:
y = 2, when
x = – 1. So,
![\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\ \mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\ \mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\ \mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\ \mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}](https://tex.z-dn.net/?f=%5Cmathsf%7B2%3D-%5C%2C%5Cdfrac%7B1%7D%7B1%5E2%2BC_1%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B2%3D-%5C%2C%5Cdfrac%7B1%7D%7B1%2BC_1%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B-%5C%2C%5Cdfrac%7B1%7D%7B2%7D%3D1%2BC_1%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B-%5C%2C%5Cdfrac%7B1%7D%7B2%7D-1%3DC_1%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B-%5C%2C%5Cdfrac%7B1%7D%7B2%7D-%5Cdfrac%7B2%7D%7B2%7D%3DC_1%7D)
![\mathsf{C_1=-\,\dfrac{3}{2}}](https://tex.z-dn.net/?f=%5Cmathsf%7BC_1%3D-%5C%2C%5Cdfrac%7B3%7D%7B2%7D%7D)
Substitute that for
C₁ into (i), and you have
![\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\ \mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\ \mathsf{y=-\,\dfrac{2}{2x^2-3}}](https://tex.z-dn.net/?f=%5Cmathsf%7By%3D-%5C%2C%5Cdfrac%7B1%7D%7Bx%5E2-%5Cfrac%7B3%7D%7B2%7D%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7By%3D-%5C%2C%5Cdfrac%7B1%7D%7Bx%5E2-%5Cfrac%7B3%7D%7B2%7D%7D%5Ccdot%20%5Cdfrac%7B2%7D%7B2%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7By%3D-%5C%2C%5Cdfrac%7B2%7D%7B2x%5E2-3%7D%7D)
So
y(– 2) is
![\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\ \mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\ \mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\ \mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}](https://tex.z-dn.net/?f=%5Cmathsf%7By%5Cbig%7C_%7Bx%3D-2%7D%3D-%5C%2C%5Cdfrac%7B2%7D%7B2%5Ccdot%20%28-2%29%5E2-3%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7By%5Cbig%7C_%7Bx%3D-2%7D%3D-%5C%2C%5Cdfrac%7B2%7D%7B2%5Ccdot%204-3%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7By%5Cbig%7C_%7Bx%3D-2%7D%3D-%5C%2C%5Cdfrac%7B2%7D%7B8-3%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7By%5Cbig%7C_%7Bx%3D-2%7D%3D-%5C%2C%5Cdfrac%7B2%7D%7B5%7D%7D%5Cquad%5Clongleftarrow%5Cquad%5Ctextsf%7Bthis%20is%20the%20answer.%7D)
I hope this helps. =)
Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>
Answer:
1(2-3)(4-7)
(1)(-1)(4-7)
-1(4-7)
(-1)(-3)
=3
Step-by-step explanation:
Answer:
22.5miles
Step-by-step explanation:
For Map 1;
1 inches = 30miles
For Map 2:
2 inches = 15miles
1 inches = x
2x = 1 *15
x = 15/2
x = 7.5miles
This means that 1 inch represents 7.5miles on Map 2'
The distance between the two cities in miles = 30miles - 7.5miles
The distance between the two cities (in miles) = 22.5miles