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3 years ago

How many solutions does the system have?20x - 5y = 54x- y = 1

Mathematics

1 answer:

irga5000 [103]3 years ago

7 0

Answer:

Infinate

Step-by-step explanation:

If you multiply all factors in the secon eqation by 5, you will get the first eqation.

Therefore for any value of x you can find y, thus having an infinate number of x,y pairs

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Find the solution to the equation below.

Andreyy89

Answer:

B and D

Step-by-step explanation:

The quadratic formula is $x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ . Here a=20, b=-26, and c=8.

Substitute and you'll have:

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a} =\frac{26+/-\sqrt{26^2-4(20)(8)} }{2(20)}=\frac{26+/-\sqrt{676-640} }{40)}$

$\frac{26+/-\sqrt{36} }{40}=\frac{26+/-6 }{40}$

This means the solutions are 26 - 6/ 40 and 26 + 6/ 40.

It simplifies to 1/2 and 4/5.

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3 years ago

The GCF of 15 and 27 is _____.

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3 years ago

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The volume of a sphere is found using the formula V= 4/3 3.14r^3. If the radius of a sphere is 3/4 feet, what is the volume of t

Rufina [12.5K]

The volume of the sphere is expressed in the formula V= 4/3 3.14r^3 and we are given the radius of the sphere. to find the volume, we can directly substitute r as the unit of the volume to be determined is consistent with the given. Hence, via subsitution, the answer is 1.7671 cubic feet

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2 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

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

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago

How many solutions does the system have?20x - 5y = 54x- y = 1​

How many solutions does the system have?20x - 5y = 54x- y = 1