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2 years ago

Find the slope of the line passing through the points (2,7) and (-1, 4).

Mathematics

2 answers:

natulia [17]2 years ago

7 0

Answer:

m= 1.

Step-by-step explanation:

To find the slope, we will need to use the slope formula:

$m= \frac{y_{2}- y_{1} }{x_{2}-x_{1}}$

Plug in the points, giving us:

$m= \frac{7-4}{2+1}$

m= 1

skelet666 [1.2K]2 years ago

4 0

Answer:

slope = 1

Step-by-step explanation:

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The table below represents the displacement of a horse from its barn as a function of time:

katovenus [111]

A. The y-intercept (b) of a linear equation is obtained when x = 0. Therefore from the given table,

y - intercept = 8

Since at time zero the displacement is 8 ft, this means that the horse was already outside the barn initially.

B. The average rate of change of the function represents the slope of the linear equation (m). This can be calculated using the formula:

average rate of change = m = (y2 – y1) / (x2 – x1)

m = (158 – 58) / (3 – 1)

m = 50

C. Since we have determine the y-intercept and the slope, we can formulate the linear equation:

y = m x + b

y = 50 x + 8

The domain is the value of x. When y = 508, x is equivalent to

508 = 50 x + 8

x = 10 hrs

6 0

2 years ago

What do I do when the output doesn't have a constant rate of change?

Solnce55 [7]

-2 gal/min

sry if it wrong

3 0

2 years ago

Which are correct representations of the inequality –3(2x – 5) < 5(2 – x)? Check all that apply.

k0ka [10]

Answer:

C. $-6x+15$

A. First Picture

Step-by-step explanation:

We are given the inequality, $-3(2x-5)$ .

On simplifying the inequality, we get,

$-3(2x-5)$

i.e. $-6x+15$

i.e. $-x$

i.e. $x>5$

So, we get the correct representations of the inequality are,

C. $-6x+15$

A. the First Picture

5 0

2 years ago

Read 2 more answers

Write the equation of the circle with center (7,-2) and radius square root 61. Please show your work!

Vladimir79 [104]

The equation of a circle is (x - h)^2 + (y - k)^2 = r^2
x and y are left alone in a circle equation, while h and k are the coordinates of the center.

Thus, our equation is (x - 7)^2 + (y + 2)^2 = 61

5 0

3 years ago

Can someone please help me with this

Mama L [17]

Answer:

27.

Equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1: 400 feet

Part 2: Velocity is +96 feet/second (or 96 feet/second UPWARD) & Speed is 96 feet per second

Part 3: acceleration at any time t is -32 feet/second squared

Part 4: t = 10 seconds

29.

Part 1: Average Rate of Change = -15

Part 2: The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

Step-by-step explanation:

27.

First of all, the equation of tangent line is given by:

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

Where m is the slope, or the derivative of the function

Now,

If we take a point x, the corresponding y point would be $x^2-3x+5$ , so the point would be $(x,x^2-3x+5)$

Also, the derivative is:

$f(x)=x^2-3x+5\\f'(x)=2x-3$

Hence, we can equate the DERIVATIVE (slope) and the slope expression through the point given (0,1) and the point we found $(x,x^2-3x+5)$

The slope is $\frac{y_2-y_1}{x_2-x_1}$

So we have:

$\frac{x^2-3x+5-1}{x-0}\\=\frac{x^2-3x+4}{x}$

Now, we equate:

$2x-3=\frac{x^2-3x+4}{x}$

We need to solve this for x. Shown below:

$2x-3=\frac{x^2-3x+4}{x}\\x(2x-3)=x^2-3x+4\\2x^2-3x=x^2-3x+4\\x^2=4\\x=-2,2$

So, this is the x values of the point of tangency. We evaluate the derivative at these 2 points, respectively.

$f'(x)=2x-3\\f'(-2)=2(-2)-3=-7\\f'(2)=2(2)-3=1$

Now, we find 2 equations of tangent lines through the point (0,1) and with respective slopes of -7 and 1. Shown below:

$y-y_1=m(x-x_1)\\y-1=-7(x-0)\\y-1=-7x\\y=-7x+1$

and

$y-y_1=m(x-x_1)\\y-1=1(x-0)\\y-1=x\\y=x+1$

So equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1:

The highest point is basically the maximum value of the position function. To get maximum, we differentiate and set it equal to 0. Let's do this:

$s(t)=160t-16t^2\\s'(t)=160-32t\\s'(t)=0\\160-32t=0\\32t=160\\t=\frac{160}{32}\\t=5$

So, at t = 5, it reaches max height. We plug in t = 5 into position equation to find max height:

$s(t)=160t-16t^2\\s(5)=160(5)-16(5^2)\\=400$

max height = 400 feet

Part 2:

Velocity is speed, but with direction.

We also know the position function differentiated, is the velocity function.

Let's first find time(s) when position is at 256 feet. So we set position function to 256 and find t:

$s(t)=160t-16t^2\\256=160t-16t^2\\16t^2-160t+256=0\\t^2-10t+16=0\\(t-2)(t-8)=0\\t=2,8$

At t = 2, the velocity is:

$s'(t)=v(t)=160-32t\\v(2)=160-32(2)\\v(2)=96$

It is going UPWARD at this point, so the velocity is +96 feet/second or 96 feet/second going UPWARD

The corresponding speed (without +, -, direction) is simply 96 feet/second

Part 3:

We know the acceleration is the differentiation of the velocity function. let's find it:

$v(t)=160-32t\\v'(t)=a(t)=-32$

hence, the acceleration at any time t is -32 feet/second squared

Part 4:

The rock hits the ground when the position is 0 (at ground). So we equate the position function, s(t), to 0 and find time when it hits the ground. Shown below:

$s(t)=160t-16t^2\\0=160t-16t^2\\16t^2-160t=0\\16t(t-10)=0\\t=0,10$