All categories

3 years ago

Y+4=-2(x-1) slope intercept form

1 answer:

6 0

To convert this equation into slope intercept form simply distribute the number of -2 and put it in

Y = mx + b form

Y + 4 = -2(X-1)
Y + 4 = -2x + 2
Y = -2X + 2 - 4

Y = -2X - 2

I believe this is the solution in slope intercept form.

You might be interested in

Question 4 options:

IgorLugansk [536]

7-k i hope thiis helps

5 0

3 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago

Find the surface area of the figure shown. 5 cm 5 cm cm 1.5 cm 8 cm your answer

Delicious77 [7]

Answer:

51 cm^2

Step-by-step explanation:

the formula for area of an icoselese triangle is
A = b * h / 2
h is a line from the angle where the 5 and 5 intersect to the middle of the third line
You can find h using the pythagoran theorm with the 5 being the hypotenuse and the a being half the third line

4^2 + b^2 = 5^2
16 + b^2 = 25
b = 3, so h = 3, so now
A = 8 * 3 / 2
A = 24 / 2
A = 12 cm^2

now we have the triangle, just multiply that by 2 (top and bottom)
12 * 2 = 24

sides
8 * 1.5 = 12
5 * 1.5 * 2 = 15

now add all of it
15 + 12 + 24 = 51

4 0

1 year ago

Determine the range of the function. (0,2) (2,4) (4,6) (6, 8) (8, 10)

Nikolay [14]

The given points, <span>(0,2) (2,4) (4,6) (6, 8) (8, 10), represent a relationship / function.

The range includes all of the y-values: {2, 4, 6, 8, 10}, which can also be expressed as [2,10]. This is equivalent to answer C.</span>

8 0

3 years ago

2(x + 1) – (–x + 5) ≤ –18.

pshichka [43]

You posted twice.

I already answered

2(x + 1) – (–x + 5) ≤ –18

2x + 2 + x - 5 <= -18

3x - 3 <= -18

3x <= -18 + 3

3x <= -15

x <= -15/3

x <= -5

3 0

3 years ago

Read 2 more answers