MATHEMATICS MIDDLE SCHOOL

Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of the function Y=2x^2+12x+13
axis of symmetry:
vertex:

Answers

Answer 1

Answer:

Answer:

Axis of symmetry: –3

Vertex: (–3, –5)

Step-by-step explanation:

For a quadratic function in standard form, the axis of symmetry is a vertical line $x=-(b)/(2a)$ . Therefore:

$x=-(b)/(2a)=-(12)/(2(2))=-(12)/(4)=-3$

For the vertex, –3 is the x-coordinate and we solve for y to find the y-coordinate:

y = 2x² + 12x + 13

y = 2(–3)² + 12(–3) + 13

y = 2(9) + 12(–3) + 13

y = 18 – 36 + 13

y = –5

Therefore, the vertex is (–3, –5).

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Please tell me the answer

Answers

This is easy. You all need to do this:
100÷2 it equals 50.
You need to subtract one 50 from either Caroline/Mohamed(Take Caroline's for example)
Caroline = 50-16=34.
Then you add 16 to 50(Mohamed's)
50+16=66
Therefore, Caroline sold 34 tickets, however, Mohamed sold 66 tickets.
To prove: add 66+34
66+34=100

Mohamed sold 58, and Caroline sold 42.

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:

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

The following data points represent the number of points scored last basketball game by each player on the Dragons basketball team.5,8,11,7,?

If the mean of the data set is 8 points, find the missing number of points.

Answers

Answer:

Step-by-step explanation:

There are 5 data points. To find the mean, we add the data and divide by 5

(5+8+11+7+?) /5 = 8

Multiply each side by 5

(5+8+11+7+?) /5 *5 = 8*5

(5+8+11+7+?) = 40

Combine like terms

? + 31 = 40

Subtract 31 from each side

? +31-31 = 40-31

? = 9

The missing number is 9

The missing number is 9.

8 1/4x5 1/2what is the answer to this

Answers

the answer is 45.375

45.375 is the answer

What is the percent of change?Initial value: 10
Final value: 60

(A) 70% (B) 500% (C) 50% (D) 600%

Answers

percent means parts out of 100
x%=x/100

find change
10 to 60=50 (60-10=50)

percent change=change/originla
original=10
change=50
50/10=x%
5/1=x/100
multiply both sides by 100
500=x

answer is 500%

B. because 100% of 10 is 10, u have 60 which is 6 times that, so 100%*6 which = 600%
but it's the change, not just that, and the change is 50, so 500%

The Sum Of The Measures Of The Three Sides Of A Triangle Is 11 cm, and the Measure of the largest side is 5 cm, and the smallest side is 2 cm. find the measure of the largest Angle.