Slope-intercept form of a line is y=mx+b.
Where m= slope and b= y-intercept.
First step is to compare the given equation y=35x+8 with the above equation to get the value of m.
After comparing the two equations we will get m=35.
Slope of paralle lines always equal which means slope of a line which is parallel to the above line will also be 35.
Now the line is passing through (-10,4).
Point slope form of a line is :
Next step is to plug in m=35, x1=-10 and y1=4 in the above equation. So,
y-4=35(x-(-10)
y-4=35(x+10)
y-4=35x+350
y=35x+350+4
y=35x+354.
So, the equation of the line is y=35x+354.
The 3 is in the ones place because it is on one right of the decimal point.
Subtracting 1 from the function notation will decrease the y-values on the graph by 1 unit.
This will shift then entire graph of f down by 1 unit to create the graph of g.
Answer:
the total she earns with "d" hours of dog walking and "b" hours of babysitting is given by: 8 d + 12 b
Step-by-step explanation:
If we use "d" to represent the number of hours that Susan walks the dog, and we define "b" as the number of hours she babysits, then the expression that represents the amount she earns by walking the dog is: $8 times d = 8 d
Similarly, the amount she earns by babysitting is $12 times b = 12 b
Then the total she earns with "d" hours of dog walking and "b" hours of babysitting is: 8 d + 12 b
Answer:
11. x = -3+√37 ≈ 3.08276
12. x = 11.2
13. x = -6 +6√5 ≈ 7.41641
Step-by-step explanation:
In each case, the relation of interest is ...
(distance to circle near) × (distance to circle far) = (distance to circle near) × (distance to circle far)
When there is only one point of intersection of the secant with the circle—because it is a tangent—then the product is the square of the length of the tangent.
11. 2(2+12) = x(x +6)
x² +6x -28 = 0
(x +3)² -37 = 0
x = -3+√37 ≈ 3.08276
12. 5(5+x) = 9(9)
5x +25 = 81
x = 56/5 = 11.2
13. x(x +12) = 12(12)
x² +12x -144 = 0
(x +6)² -180 = 0
x = -6 +√180 ≈ 7.41641
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<em>Comment on this secant rule</em>
This rule turns out to apply whether the point of intersection of the secant lines is outside the circle (as in these problems) or inside the circle (as in problem 9). The product of the two distances from the point of intersection to the circle is a constant for a given pair of intersecting secants/chords.
