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3 years ago

Simplify (3x − 5) − (4x + 6).

Mathematics

1 answer:

irinina [24]3 years ago

3 0

(3x − 5) − (4x + 6)

Pretend that there is a 1 in front of the first bracket.

Pretend that there is a -1 in front of the second bracket.

Multiply the first bracket by 1

(1)(3x)=3x, (1)(-5)=-5

Multiply the second bracket by -1

(-1)(4)=-4, (-1)(6)=-6

3x-5-4x-6

3x-4x-5-6 ( combine like terms)

-x-11

Answer: -x-11 or -11-x

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Check all of the ordered pairs that satisfy the equation below.y=2/5x

zvonat [6]

We are going to do this step by step:

Let's start with option A

X = 25 and Y = 5/2

$y=\frac{2}{5}\cdot x=\frac{2}{5}\cdot25=\frac{2\cdot25}{5}=\frac{50}{5}=\frac{5\cdot10}{5\cdot1}=10$

In this case, when X = 25 , then Y = 10 ,which is different to 5/2

X = 14 , Y = 35

$y=\frac{2}{5}\cdot14=\frac{28}{5}$

In case, when X = 14, then Y = 28/5, which is different to 35

X = 40 , Y = 24

$y=\frac{2}{5}\cdot40=\frac{80}{5}=16$

Similarly, in this case, when X = 40, then Y = 16, which is different to 24

X = 10 , Y=4

$y=\frac{2}{5}\cdot10=\frac{20}{5}=4$

Now, in this case, we can see that when X = 10, then Y = 4 which is the same as the given value of Y

X = 50 , Y = 20

$y=\frac{2}{5}\cdot50=\frac{100}{5}=20$

In this case, the values of Y are also the same.

X = 30 , Y = 12

$y=\frac{2}{5}\cdot30=\frac{60}{5}=12$

Again, in this case, the values of Y are the same, so the pair satisfies the equation.

In conclusion: options D, E and F satisfy the equation

3 0

1 year ago

1. Miles and Simon do chores to earn an allowance from their parents. Miles has $120 saved and gets a $10 allowance per week. Hi

Ket [755]

Answer:

If w represented the number of weeks he does his chores, the equations would be:

Miles: 120+10w

Simon:75+15w

7 0

3 years ago

Locate the prepositional phrase and type it in the blank. Then, identify the function of the phrase. Above the door hung a horse

olga_2 [115]

“Above the door” is the prepositional phrase but I can’t help you you it’s the second part, sorry.

3 0

3 years ago

Suppose that a large mixing tank initially holds 100 gallons of water in which 50 pounds of salt have been dissolved. Another br

Serggg [28]

Answer:

dA/dt = 12 - 2A/(100 + t)

Step-by-step explanation:

The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 4 lb/gal

Input rate of brine = 3 gal/min

Thus;

R_in = 4 × 3 = 12 lb/min

Due to the fact that solution is pumped out at a slower rate, thus it is accumulating at the rate of (3 - 2)gal/min = 1 gal/min

So, after t minutes, there will be (100 + t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(100 + t)]lb/gal × 2 gal/min

R_out = 2A(t)/(100 + t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 12 - 2A(t)/(100 + t)

Since we are to use A foe A(t), thus the Differential equation is now;

dA/dt = 12 - 2A/(100 + t)

5 0

3 years ago

When rounding to the nearest hundred i become 500 what numbers could i be

RUDIKE [14]

Any number above 450

4 0

3 years ago

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