Simplify -6 + 4/7 x + 4 - 6/7 x , show work

Please I Need Help Asap .

Answers

Answer 1

Answer: 6+4= 10
4/7x+-6/7x= -2/7x
Answer would be 10 - 2/7x

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2.789 in expanded form

What is 9\4 as a mixed number

Answers

Okay so the formula is

1. Divide the numerator by the denominator.
2. Write down the whole number answer.
3. Then write down any remainder above the denominator.

so your answer is 2 1/4

U can divide 9 by 4 which equals 2.25 in decimal form but in fraction from it would look like this 2 1/4

(1 pt) Which graphs show the solution to the given inequalities? Column A Column B 1. 2. 3. 4. A. Number line with positive and negative numbers. A ray starts at negative 12 with an open circle and extends to the right. B. Number line with positive and negative numbers. A ray starts at negative 12 with a solid circle and extends to the right. C. Number line with positive and negative numbers. A ray starts at negative 12 with a solid circle and extends to the left. D. Number line with positive and negative numbers. A ray starts at negative 12 with an open circle and extends to the left.

Answers

I don't know but its at the tip of my tounge

Rhea is solving a math puzzle. To find the solution of the puzzle, she must find the product of two numbers. The first number is the sum of 23 and x, and the second number is 18 less than two times the first number. Which of the following functions represents the product of these two numbers?Opions are:

A.
P(x) = 2x^2 + 74x + 644
B.
P(x) = 2x^2 + 28x - 414
C.
P(x) = 2x^2 + 102x + 1,288
D.
P(x) = 2x^2 + 110x + 1,472

Answers

Answer:

It's B.

Step-by-step explanation:

First is x + 23 and second is 2x - 18.

So the product is

(x + 23)(2x - 18)

= 2x^2 - 18x + 46x - 414

= 2x^2 + 28x - 414

A bus goes from town A to B in an exact time. If the bus goes at the rate of 50km/h, then it will arrive B 42min later than it had to and if it increases its speed 5.5/9 m/sec., it will arrive B 30min earlier. Find:A) The distance between the two towns;
B) The exact time that it takes to arrive town B
C) The speed of the bus(by schedule) for the exact time.

Answers

Let the speed of bus for the exact time = x km/h
the distance between the cities = y km
then the exact time would be, t hours = (y/x) hours

If the bus goes at the rate of 50km/h, then it will arrive B 42min later,
speed = 50 km/h
42 minutes = 42/60 hours = 7/10 hours = 0.7 hours
time taken = t+0.7
distance = speed×time
⇒ y = 50×(t+0.7)
⇒ y = 50t + 35 ---------------------(1)

it increases its speed 5.5/9 m/sec, it will arrive B 30min earlier.
5.5/9 m/s = (5.5/9)×(18/5) km/h = 2.2 km/h
30 minutes = 30/60 = 0.5 hour
speed = (x+2.2) km/h
time = (t - 0.5) hours

distance = speed×time
⇒ y = (x+2.2)×(t-0.5)
⇒ y = ((y/t) +2.2)×(t-0.5) (t = y/x)
⇒ y = y - 0.5 (y/t) + 2.2t - 1.1
⇒ 0.5 (y/t) - 2.2t + 1.1 = 0   (subtracting y from both sides)
⇒ (y/t) - 4.4t - 2.2 = 0    (dividing both sides by 0.5)
⇒ y - 4.4t² - 2.2t = 0    (multiplying both sides by t)
⇒ 50t + 35 - 4.4t² - 2.2 t = 0 (from equation 1)
⇒ -4.4t² + 35 + 47.8t = 0
⇒ 4.4t² - 47.8t - 35 = 0

solving the quadratic equation, we get t = 11.55 hours
y = 50t + 35 = 612.5 km
x = 612.5/11.55 = 53 km/h

A) 612.5 km
B) 11.55 hours
C) 53 km/h

The Logic Defined:

1 Minute=t, (a unit of time)

Time (By schedule)=nt, (n>0), nt=number of minutes

Metre(s)=m

Speed=s (in metres per minute), s=[distance in metres]/[time in minutes]

Distance=d (in metres), d=[speed in metres per minute]*[time in minutes]

---------------------------------------------

Statement (1):

"If the bus goes at the rate of 50km/h, then it will arrive B 42min later than it had to."

Conclusion 1:

$\frac { 50km }{ h } =\frac { 50,000m }{ 60t } =\frac { 2,500m }{ 3t } \n \n \therefore \quad \frac { 2,500m }{ 3t } =nt+42t\n \n \frac { 2,500m }{ 3t } =t\left( n+42 \right)$

$\n \n 2,500m=3{ t }^( 2 )\left( n+42 \right) \n \n m=\frac { 3{ t }^( 2 )\left( n+42 \right) }{ 2,500 }$

Statement (2):

"if it increases its speed 5.5/9 m/sec., it will arrive B 30min earlier."

Conclusion 2:

$\frac { 5.5m }{ 9\quad seconds } =\frac { 5.5m }{ \frac { 9 }{ 60 } t } =\frac { 110m }{ 3t }$

$\n \n \therefore \quad \frac { 110m }{ 3t } =nt-30t\n \n \frac { 110m }{ 3t } =t\left( n-30 \right) \n \n 110m=3{ t }^( 2 )\left( n-30 \right)$

$\n \n m=\frac { 3{ t }^( 2 )\left( n-30 \right) }{ 110 }$

Conclusion 3, because of conclusion 1 and 2:

$\frac { 3{ t }^( 2 )\left( n-30 \right) }{ 110 } =\frac { 3{ t }^( 2 )\left( n+42 \right) }{ 2,500 } \n \n 7,500{ t }^( 2 )\left( n-30 \right) =330{ t }^( 2 )\left( n+42 \right) \n \n 7,500\left( n-30 \right) =330\left( n+42 \right)$

$\n \n 7,500n-225,000=330n+13,860\n \n 7,500n-330n=13,860+225,000\n \n 7,170n=238,860\n \n n=\frac { 238,860 }{ 7,170 } \n \n \therefore \quad n=\frac { 7962 }{ 239 }$

Therefore,

$Time\quad by\quad schedule=\frac { 7962 }{ 239 } t\n \n Approx:\quad 33.3\quad mins$

Now we want to find the distance between the two towns, so we say that:

$d=\frac { 2,500m }{ 3t } \cdot \left( \frac { 7962 }{ 239 } t+42t \right) \n \n =\frac { 2,500m }{ 3t } \cdot \frac { 18,000 }{ 239 } t$

$\n \n =\frac { 45,000,000 }{ 717 } m\n \n Approx:\quad 62,761.5\quad metres\n \n In\quad km\quad (approx):\quad 62.761\quad km$

So now you want to know how fast the bus has to travel to get to its destination on time...

Use the formula: s=d/t

Therefore:

$s=\frac { \frac { 45,000,000 }{ 717 } m }{ \frac { 7962 }{ 239 } t } \n \n Approx:\quad 1,883.9\quad metres\quad per\quad minute$

Janine has job offer sat two companies. One company offers a starting salary of 28,000 with a raise of 3,000 each year. The other company offers a starting salary 36,000 with a raise of 2,000 each year A. After how many years would Janine's salary be the same with both companies

B. what would the salary be ?

Show work please

Answers

28,000 +3,000= 31,000 +3,000 = 34,000 +3,000 = 37,000 +3,000 = 40,000 +3,000 = 43,000 + 3,000 = 46,000

36,000 +2,000= 38,000 +2,000 = 40,000 +2,000 = 42,000 +2,000 = 44,000 +2,000 = 46,000

THE SALARY WOULD BE 46,000
IT WOULD TAKE 6 YEARS

I hope i did that right

Please help me I'm on a time crunch