Which situation could be solved using the equation-4+4=0

The first picture is the graph (already answered) and the second is the next

Nastasia [14]

Answer:

a) 100/15 = 6.6667 = 6 2/3

b) C = T( 6 2/3)

= T (6 + 2/3)

C = 1T x 6 2/3

= 6 2/3 T

We can show T as time at end of equation or start of brackets

8 0

2 years ago

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What is the equation of the line that passes through 1,4 and 2,1

sashaice [31]

Answer:

y = -3x + 7

Step-by-step explanation:

The equation of a line

y = mx + c

y - intercept point y

m - slope of the line

x - intercept point x

c - intercept point of the line

Step 1: find the slope

m = y2 - y1 / x2 - x1

Given two points

( 1 , 4) ( 2 , 1)

x1 = 1

y1 = 4

x2 = 2

y2 = 1

insert the values

m = 1 - 4 / 2 - 1

m = -3/1

m = -3

y = -3x + c

Step 2: substitute any of the two points given into the equation of a line

y = -3x + c

( 1 ,4)

x = 1

y = 4

4 = -3(1) + c

4 = -3 + c

4 + 3 = c

c = 7

Step 3: sub c into the equation

y = -3x + 7

The equation of the line is

y = -3x + 7

3 0

3 years ago

The force needed to stop a car varies directly as its weight W and the square root of velocity V are inversely as the distance d

svet-max [94.6K]

Answer:

Step-by-step explanation:

8 0

2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

2 years ago

Hi, I have utterly no idea what I'm doing here.

Anna11 [10]

Answer:

Factored form: 2x(2x^2+x+1)

Step-by-step explanation:

You're subtracting them. :)

6 0

2 years ago

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