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2 years ago

Solve the systems of equations 6x+3y=9 2x+3y=1

1 answer:

7 0

Solve by elimination.
The goal is to cancel out one of the variables in order to easily solve for the other variable.

Do this by changing the equations so that the coefficients of either x or y add up to 0.

Notice the coefficients of y are 3 and 3, if we make one of them negative then they add up to 0. 3+ (-3) = 0

Multiply 2nd equation by -1.

6x +3y = 9
-2x -3y = -1
__________
4x +0y = 8

Solve for x
4x = 8
x = 8/4 = 2

Plug x=2 back into one of original equations to find y.
---> 2(2) + 3y = 1
---> 4 + 3y = 1
---> 3y = -3
---> y = -1

Therefore solution is (2,-1)

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Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

Joe is considering taking out along with the principal of $15,500 From one of two banks. Bank A charges and interest rate of 6%,

ki77a [65]

Answer:

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total cost of the loan at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount taken as loan.

Considering bank A's offer

From the information given,

P = $15500

r = 6% = 6/100 = 0.06

n = 12 because it was compounded 12 times in a year.

t = 8 years

Therefore,

A = 15500(1 + 0.06/12)^12 × 8

A = 15500(1 + 0.005)^96

A = 15500(1.005)^96

A = $25019.2

The interest paid is

25019.2 - 15500 = $9519.2

Considering bank B's offer

From the information given,

P = $15500

r = 6.5% = 6.5/100 = 0.065

n = 12 because it was compounded 12 times in a year.

t = 7 years

Therefore,

A = 15500(1 + 0.065/12)^12 × 7

A = 15500(1 + 0.0054)^84

A = 15500(1.0054)^84

A = $24366

The interest paid is

24366 - 15500 = $8866

The interest that would by paid on bank B's offer is lower than that of bank A and the duration of the loan offer from bank B is shorter than that of bank A.

Therefore,

Bank A gives you more time to pay up the loan and you end up paying more interest

Bank gives you lesser time to pay up the loan and you end up paying lesser interest.

3 0

3 years ago

which of the following formulas would find the lateral area of a right cylinder where h is the height and r is the radius

vlada-n [284]

Answer:

B. $LA=2\pi rh$

Step-by-step explanation:

The lateral area of the right cylinder refers to the curved surface area.

The lateral area of the right cylinder does not include the two circular bases.

The lateral area is given by the formula;

$LA=2\pi rh$

The correct choice is B.

6 0

2 years ago

Read 2 more answers

In triangle STU, u2 = s2 + t2. Triangle STU has sides s, t, u opposite to the corresponding vertices S, T, U Which equation is t

wlad13 [49]

Answer:
<span>The measure of angle SUT is equal to 90 degrees
</span>
Explanation:
The Pythagorean theorem states that:
"In any right-angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides"
In other words:
(hypotenuse)² = (first side)² + (second side)²

Now, for the given, we have:
u² = s² + t²
This means that the given triangle UST is a right-angled triangle and that side "u" is its hypotenuse.

According to the given, the angle opposite to side "u" is angle U. This means that angle U is the right-angle in this triangle.

Based on the above:
angle U = 90°
angle S + angle T = 180 - 90 = 90°

Now, comparing our deductions to the given choices, we can conclude that the correct choice is:
The measure of angle SUT is equal to 90 degrees

Hope this helps :)

6 0

3 years ago

Read 2 more answers

Given b(x) = X+41, what is b(-10)?<br> О -10<br> боооо

ch4aika [34]

Answer: 31
step by step solution: so the function is b(x)=x+41. they want to know what the function would equal when b(-10). so b(-10) means that you would plug (-10) in for x in the function so the function would now be b(x)=(-10)+41. now all you have to do is simplify the function. b(x)=31. so the function b(-10)=x+41 equals 31. hope this helped!!! have a great day.

6 0

2 years ago