MATHEMATICS MIDDLE SCHOOL

How I solve x*x
Please help
1) 2x
2) x^2
3) x(x+2)
4) x+2

Answers

Answer 1
Answer: It would be 2) x^2.

In this question, if you are having trouble, we could replace x with a number.

For example, x=5. Then, it would be 5 × 5 or 5^2.

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If every positive number has two square roots and you can find the length of the side of a square window by finding a square root of the area, why isthere only one answer for the length of a side?

Answers

because the 2 square roots are positive and negative
ex
squareroot of 4=2 or -2 since 2 time2=4 and -2 times -2=4
we normally use the principal root or the positive root since legnth in real life cannot be negative

you cannot have a window that is -5 inches long

th enaswe ris because one s negative and cannot be used

-5(2w+1)=25 how do I solve this problem

Answers

Answer:

Step-by-step explanation:

Divide by - 5

-5(2w + 1) / - 5 = 25 / - 5

2w + 1 = - 5              

Subtract 1 from both sides.

2w + 1 - 1 = - 5 - 1

2w = - 6

Divide by 2

2w/2 = -6/2

w = - 3

-5(2w+1)=25

Multiply both sides by -5

(-5)(2w)=-10w

(-5)(1)=-5

-10w-5=25

Move -5 to the other side. Sign changes from -5 to +5.

-10w-5+5=25+5

-10w=30

Divide by -10 for both sides

-10w/-10=30/-10

w=-3

Answer: w=-3

How to divide 8.4 into 0.03

Answers


8.4 / 0.03 would be 280.
Hope this helps! :)
take out the points and then divide .we  will get exact answer as 280

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

Tammy is going for a walk she walks at a speed of 3 miles per hour for 7.5 miles. For how many hours does she walk.

Answers

Tammy does walk a time of  3.75 hours

Explanation:

Given-

Speed, (which can be represented as  s) = 2 miles/hour

Distance, (which can be represented as d) = 7.5 miles of distance .

Time, t = ?

We know,

d = s t

7.5 miles of distance  = 2 miles/hour × t

3.75 hour of time  = t

Therefore, Tammy does walk a time of  3.75 hours

HELP ME PLEASE I BEG U1!!!!!

Answers

the first one is 41, since it occurs most frequently in the data
second one is 5.6, since it is in the middle of the data
working on others.....
The answer of the first question is 41 because the majority of the number in the list is 41 and there are three of them.

The answer of the second question is 3.4 which is in middle part in the list.
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