Please help me what is this

Answers

Answer 1

Answer: 36 ÷ 9 = 4
So.. 1% is 4 people
23 + 47 = 70%
70 × 4 = 280
Hope this helped :)

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Find the value of B - A if the graph of Ax + By = 3 passes through the point (-7, 2), and is parallel to the graph x + 3y = -5. Pls help ASAP btw, the answer isn’t -12/19 or 12/19

Answers

Answer:

$B - A = -6$

Step-by-step explanation:

Given

Point: (-7,2)

x + 3y = -5

Required

Find B- - A in Ax + By = 3

To start with; we need to calculate the slope of x + 3y = -5

$x + 3y = -5$

Subtract x from both sides

$x - x + 3y = -5 - x$

$3y = -5 - x$

Divide both sides by 3

$(3y)/(3) = -(5)/(3) - (x)/(3)$

$y = -(5)/(3) - (x)/(3)$

The slope of the line is the coefficient of x

Slope = $- (1)/(3)$

The question says line Ax + By = 3 is parallel to line x + 3y = -5; This means that they have the same slope of $- (1)/(3)$

Having calculated the slope, next is to calculate the equation of the line using the following formula;

$m = (y - y_1)/(x - x_1)$

Where m is the slope; m = $- (1)/(3)$ ; $(x_1, y_1) = (-7,2)$

Substitute these values in the formula above; the formula becomes

$-(1)/(3) = (y - 2)/(x - -7)$

$-(1)/(3) = (y - 2)/(x +7)$

Cross Multiply

$-1(x+7) = 3(y-2)$

Open brackets

$-x - 7 = 3y - 6$

Add x to both sides

$x - x - 7 = 3y - 6 + x$

$-7 = 3y - 6 +x$

Add 6 to both sides

$-7 + 6 = 3y -6 + 6 + x$

$-1 = 3y + x$

Multipby both sides by -3

$-3(-1) = -3(3y + x)$

$3 = -9y - 3x$

$-9y - 3x = 3$

$-3x - 9y = 3$

Comparing the above to Ax + By = 3

$Ax = -3x\nA = -3$

$By = -9y\nB = -9$

$B - A = -9 - (-3)$

$B - A = -9 + 3$

$B - A = -6$

What fraction of 1 gallon is 5 cups

Answers

5/16.
4 quarts to a gallon. 4 cups to a quart. each cup is 1/16 of a gallon. therefor 5/16

0.2(d − 6) = 0.3d + 5 − 3 + 0.1d. Help needed

Answers

$\text{Solve for d}\n\n0.2(d − 6) = 0.3d + 5 − 3 + 0.1d\n\n\text{Combine like terms:}\n\n0.2(d - 6) = 0.4d + 2\n\n\text{Use the distributive property:}\n\n0.2(d - 6) = 0.4d + 2\n\n0.2d-1.2=0.4d+2\n\n\text{Subtract 0.4d from both sides}\n\n-0.2d-1.2=2\n\n\text{Add 1.2 to both sides}\n\n-0.2d=3.2\n\n\text{Divide both sides by -0.2}\n\n\boxed{d=-16}$

Sequences kind of confuse me, so could someone help?

Answers

Answer:

Step-by-step explanation:

Consider the number of triangles in each diagram

1 → 1 → $2^(0)$

2 → 2 → $2^(1)$

3 → 4 → 2²

4 → 8 → 2³

The pattern of triangles is 1, 2, 4, 8, that is

$2^(0)$ , $2^(1)$ , 2², 2³

Note that the exponent is one less than the figures position in the pattern, thus

$a_(n)$ = $2^(n-1)$ → C

A plumber charges $50 to visit a house plus $40 for every hour of work. Define a variable and write an expression to represent the total cost of hiring a plumber

Answers

The expression for the total cost of hiring a plumber will be equal to $50 + $40(h).

What is an expression?

Mathematical actions are called expressions if they have at least two terms that are related by an operator and include either numbers, variables, or both. Adding, subtraction, multiplying, and division are all reflection coefficient operations. A mathematical operation such as reduction, addition, multiplication, or division is used to integrate terms into an expression.

As per the given statement in the question,

The charge for a plumber to visit the house = $50

The additional charge per hour = $40

The expression for t hours will be,

A = $50 + $40(h),

Here, A is the total amount and h is the number of hours the plumber is working.

To know more about the expression:

brainly.com/question/14083225

#SPJ2

x = # of hours worked

Total cost = 40x + 50

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:

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

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