Hello!
The problem has asked that we write a
point-slope
equation of the line in the image above.
Point-Slope Form uses the following formula:
y –

= m(x –

)
In this case, M represents the
slope while

and

represent the
corresponding X and Y values of any given point on the line.
We are given that the slope of the line is -

. We also know that any given point on a graph takes the form (x,y). Based on the single point provided in the image above, we can determine that

is equal to
6 and

is equal to
2. Now insert all known values into the point-slope formula above:
y – 2 = -

(x – 6)
We have now successfully created an equation based on the information given in the problem above. Looking at the four possible options, we can now come to the conclusion that
the answer is C.
I hope this helps!
Answer: there are 8 quarters in the jar
Step-by-step explanation:
Let x represent the number of nickels that is inside the jar.
The total number of coins inside the jar is 20.
Probability is expressed as
Number of possible or favorable outcomes/total number of outcomes.
The probability of selecting a nickel is 0.6. This is expressed as
0.6 = x/20
Cross multiplying, it becomes
x = 20 × 0.6
x = 12
Since there are 12 nickels inside the jar, then the number of quarters would be
20 - 12 = 8
Answer:
y=-50x-20
Step-by-step explanation:
Answer:
The line segments AB and CD are perpendicular to each other.
Step-by-step explanation:
Segment AB falls on line 2x − 4y = 8.
Rearranging the equation into slope-intercept form we get, ............. (1)
Therefore, slope of the line segment AB is
Now, the segment CD falls on line 4x + 2y = 8.
Rearranging the equation into slope-intercept form we get, ............. (2)
Therefore, the slope of the line segment CD is, N = - 2
So, M × N =
Hence, we can conclude that the line segments AB and CD are perpendicular to each other
Finding the upper and lower bounds for a definite integral without an equation is pretty hard because how can we find the upper and lower bounds of definite integral if there is no equation given. But I will teach you how to find the lower and upper bounds of a definite integral, when the equation is like this
So, i integrate this,

I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4