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2 years ago

22= 18x + 4 please help!!

Mathematics

2 answers:

vredina [299]2 years ago

7 0

Answer:

x=1

Step-by-step explanation:

22-18=4 18-18=/

18x1=18

22=18(1)+4

22-18(1)=4

stellarik [79]2 years ago

3 0

Answer:

22=18x + 4

Step-by-step explanation:

collect like terms 22 and +4

+4 crosses equality sign and turns to -4

you get 22-4=18x

therefore 18x=18

cancel 18 in both sides x = 1

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Evaluate x - y - (-z) if x = -3, y = -7, and z = -9

IceJOKER [234]

It would be -6 x+9 i think.

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2 years ago

Use substitution. What is the solution to the system of equations? Use the drop-down menus to explain your answer.

Hoochie [10]

Answer:

The system of equations has one solution. The two equations are intersecting lines.

Step-by-step explanation:

if you want to use the substitution method, you would have to plug in y for 12x+2 in the second equation.

2(12x+2)=x+4

next, you have to distribute

24x+4=x+4

subtract 24x from each side

4=-23x+4

subtract 4 from each side

0=23x

divide each side by 23

x=0

now, plug in 0 to one equation.

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7 0

2 years ago

WILL GIVE BRAINLIST AS SOON AS I CAN i thought that this one was 17157.3 but its not. i feel like im close, though.

vladimir2022 [97]

Answer:

2143.6 ft³

Step-by-step explanation:

We use formula number 2 to get the volume. V=4\3πr³

4\3 x 3.14 x 8³ = 2143.573333

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4 0

2 years ago

The manufacturer of a CD player has found that the revenue R (in dollars) is Upper R (p )equals negative 5 p squared plus 1 com

AleksAgata [21]

Answer:

The maximum revenue is $1,20,125 that occurs when the unit price is $155.

Step-by-step explanation:

The revenue function is given as:

$R(p) = -5p^2 + 1550p$

where p is unit price in dollars.

First, we differentiate R(p) with respect to p, to get,

$\dfrac{d(R(p))}{dp} = \dfrac{d(-5p^2 + 1550p)}{dp} = -10p + 1550$

Equating the first derivative to zero, we get,

$\dfrac{d(R(p))}{dp} = 0\\\\-10p + 1550 = 0\\\\p = \dfrac{-1550}{-10} = 155$

Again differentiation R(p), with respect to p, we get,

$\dfrac{d^2(R(p))}{dp^2} = -10$

At p = 155

$\dfrac{d^2(R(p))}{dp^2} < 0$

Thus by double derivative test, maxima occurs at p = 155 for R(p).

Thus, maximum revenue occurs when p = $155.

Maximum revenue

$R(155) = -5(155)^2 + 1550(155) = 120125$

Thus, maximum revenue is $120125 that occurs when the unit price is $155.

6 0

3 years ago

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ArbitrLikvidat [17]

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6 0

3 years ago