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harina [27]
2 years ago
15

What’s the answer?????

Mathematics
2 answers:
MAXImum [283]2 years ago
4 0
1st simplify
8x+32=104
2nd subtract 32 from both sides
8x+32=104
-32 -32
3rd simplify
8x=72
4th solve for x
8x=72
divide both sides by 8
x=9
posledela2 years ago
3 0

Answer:9

Step-by-step explanation: x+4=13

x=13-4

x=9

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43.7 % is equivalent to 0.437 .
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If the Federal Reserve sets the reserve rate to 2%, what is the resulting money
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(no links please and than you! :))
pashok25 [27]

Answer:

Line D represents a proportional relationship.

Step-by-step explanation:

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Solve for x & y solve x = 3 + yy = -2x + 9
sergey [27]

x = 3 + y Eqn(1)

y = -2x + 9​ Eqn(2)

Let us solve the system of equations with the substitution method

x - 3 = y (Subtracting 3 from both sides of the Eqn(1))

Replacing y = x - 3 in Eqn (2), we have:

x - 3 = -2x + 9

x = -2x + 9 + 3 (Adding 3 to both sides of the equation)

x + 2x = 9 + 3 (Adding 2x to both sides of the equation)

3x = 12 ( Adding like terms)

x = 12/3 (Dividing by 3 on both sides of the equation)

x = 4

Replacing x=4 in Eqn(1), we have:

4 = 3 + y

4 - 3 = y (Subtracting 3 from both sides of the equation)

y=1

The answers are:

x= 4 and y=1

6 0
1 year 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

8 0
3 years ago
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