ONE GEOMETRY QUESTION

The amount of money,y, in cash box after x tickets are purchased for carnival games. The slope of the line is 1/4 and the y-inte

victus00 [196]

Amount of money received when x tickets are sold=$Y

Number of tickets sold=X

Slope of line= $\frac{1}{4}$

And , y- intercept = 8

Equation of line having slope m and y-intercept c is given by,

y = m x + c,

Expressing the above statement in terms of linear equation in two variable i.e a line is

Y = $\frac{1}{4}$ X + 8

5 0

3 years ago

The task is in the attachment, thank you

Novosadov [1.4K]

Answer:

m=0

Step-by-step explanation:

mx²+2x-1=0

if x=1/2 then

m(1/2)² +2(1/2)-1=0

m/4+1-1=0

m/4=0

m=0

4 0

3 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

Hence, the amount of oil was decreasing at 69300 barrels, yearly

7 0

2 years ago

Please help me thank you!

sergiy2304 [10]

Answer:

The equation of the line that passes through the points

(5,2) and (-5,6)

is

y=-2/5x+4

Step-by-step explanation:

You want to find the equation for a line that passes through the two points:

(5,2) and (-5,6).

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

First, let's find what m is, the slope of the line...

The slope of a line is a measure of how fast the line "goes up" or "goes down". A large slope means the line goes up or down really fast (a very steep line). Small slopes means the line isn't very steep. A slope of zero means the line has no steepness at all; it is perfectly horizontal.

For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:

So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (5,2), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=5 and y1=2.

Also, let's call the second point you gave, (-5,6), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=-5 and y2=6.

Now, just plug the numbers into the formula for m above, like this:

m= 6 - 2/-5 - 5 or m= 4-10 or m=-2/5

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=-2/5x+b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(5,2). When x of the line is 5, y of the line must be 2.

(-5,6). When x of the line is -5, y of the line must be 6.

Because you said the line passes through each one of these two points, right?

Now, look at our line's equation so far: y=-2/5x+b. b is what we want, the -2/5 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (5,2) and (-5,6).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!.

You can use either (x,y) point you want..the answer will be the same:

(5,2). y=mx+b or 2=-2/5 × 5+b, or solving for b: b=2-(-2/5)(5). b=4.

(-5,6). y=mx+b or 6=-2/5 × -5+b, or solving for b: b=6-(-2/5)(-5). b=4.

3 0

2 years ago

Find inverse of y=1/2x+3

yKpoI14uk [10]

Answer: f⁻¹(x)=2x-6

Step-by-step explanation:

To find the inverse of an equation, you replace y with x and x with y. Then, you solve for y.

y=1/2x+3 [replace y with x and x with y]

x=1/2y+3 [subtract both sides by 3]

x-3=1/2y [multiply both sides by 2]

y=2(x-3) [distribute]

y=2x-6

Now, we know the inverse function is f⁻¹(x)=2x-6.

8 0

3 years ago