Answer:
0.2
Step-by-step explanation:
Multiply 4/5 and 1/4: Decimal form
.8 • .25
The value of "b" is the y-intercept.
In order to figure out slope-intercept form you need 1 coordinate and the slope.
1) Find the slope, using the 2-point slope formula: "m= y2-y1 / x2-x1".
ex. m= -5 - 3 / -4 - -6 (simplify)---> m= -4
2) Fill in the blanks for point-slope formula: "y - y1 = m (x - x1)"
(choose one coordinate, it doesn't matter which one)
ex. y - -5 = -4 (x - -4)
3) Then use basic algebra to simplify.
Answer:
D. 10x - 5
General Formulas and Concepts:
<u>Algebra I</u>
- Terms/Coefficients/Degrees
Step-by-step explanation:
<u>Step 1: Define</u>
f(x) = 2x + 5x
g(x) = 3x - 5
(f + g)(x) is f(x) + g(x)
<u>Step 2: Simplify</u>
f(x) = 7x
g(x) = 3x - 5
<u>Step 3: Find</u>
- Substitute: (f + g)(x) = 7x + (3x - 5)
- Combine like terms: (f + g)(x) = 10x - 5
Pretty sure the answer is D:
Because the equation is in the format y=mx+c (c being the y intercept), we know that the y intercept (or where the line will cross the y axis) will be -9.
B also says -9 but we have to be careful because it states that the x axis is -9 rather than the y axis.
Hope that helped you understand.
It's not obvious here, but you're being asked to find a linear equation for the velocity of the car, given two points on the line that represents this velocity.
Find the slope of the line segment that connects the points (3 hr, 51 km/hr) and (5 hr, 59 km/hr). Graph this line. Where does this line intersect the y-axis? Find the y-value; it's your "y-intercept," b.
Now write the equation: velocity = (slope of line)*t + b
The units of measurement of "slope of line" must be "km per hour squared," and those of the "y-intercept" must be "km per hour."
Part B: Start with the y-intercept (calculated above). Plot it on the vertical axis of your graph. Now label the horizontal axis in hours: {0, 1, 2, 3, 4, 5, 6}. Draw a vertical line through t=6 hours. It will intercept both the horiz. axis and the sloping line representing the velocity as a function of time. Show only the part of the graph that extends from t=0 hours to t=6 hours.
