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3 years ago

Solve the system algebraically. Check your work.

Mathematics

2 answers:

Mumz [18]3 years ago

8 0

x=2,y=0

plug it in and check.

Alchen [17]3 years ago

4 0

Answer:

x=2 and y=0 is the required result.

Step-by-step explanation:

We have been given system of equations:

$5x+2y=10$ (1)

And $3x+2y=6$ (2)

We will use elimination method:

Multiply 1st equation by 3 and 2nd equation by 5 we get:

$15x+6y=30$ (3)

$15x+10y=30$ (4)

Now subtract (4) from (3) we get:

$-4y=0$

$y=0$

Now, put y=0 in (1) equation:

$5x+2(0)=10$

$5x=10$

$x=2$

Hence, x=2 and y=0

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nasty-shy [4]

Answer:

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2 years ago

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is

Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = $\frac{dx}{dt}$

ship B is sailing north at 20 km/h = $\frac{dy}{dt}$

here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find $\frac{dz}{dt}$ at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

$z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\$

$z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h$

derivative first equation w . r. to t we get

$2z\frac{dz}{dt} =$ $-2(130-x)\frac{dx}{dt}$ $+2y\frac{dy}{dt}$

$\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]$

$\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }$

$= \frac{1100}{85.44}\\ = 12.87km/h$

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2 years ago

Point A(3,2) is on a circle whose center is C(−2,3). What is the radius of the circle?

goldfiish [28.3K]

Answer:

The radius of the circle is √26.

Step-by-step explanation:

Since point A is on the circle, the distance between that point and the center of the circle is equal to the radius.

To find the distance between two points, the following equation should be used:

$x^{2} =(x_{A}- x_{C}) ^{2} + (y_{A}- y_{C}) ^{2}$

Replacing the numeric values:

$r^{2} =(3- (-2)) ^{2} + (2-3) ^{2}\\r^{2} =(5) ^{2} + (-1) ^{2}\\r=\sqrt{26}$

The radius of the circle is √26.

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3 years ago

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2 years ago

Mnenie [13.5K]

The answer would have to be B

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3 years ago