Answer:
y = 5/7 x - 6
Step-by-step explanation:
For two lines to be perpendicular, the product of their slope must be -1
The slope of the given function is -7/5
Let the slope of the required function be m
m * -7/5 = -1
7/5 m = 1
7 = 5m
m = 5/7
The slope of the line perpendicular to the line is 5/7
The required equation (using any value of the y-intercept) is expressed as;
y = 5/7 x - 6
Note that the y-intercept value was assumed and any value can be used
Answer:
a. domain: {0, 2, 4}, range: {2, 6, 10}
Step-by-step explanation:
The given diagram depicts a function.
Let us define function:
A function is a mapping of inputs to output.
The left ones are the input of the function while the right side values are the output of the function.
The domain of the function is the set of values that are given as input to the function while the outputs are called range.
so, in the given question
The domain is: {0,2,4} and the range is: {2,6,10}
So, Option A is correct ..
Answer:
y=40
Step-by-step explanation:
The formula for Direct Variation is y=kx or k=y/x. In this case I would use k=y/x. If you're y is 8 and x is 3, this means that K=8/3. Using this, we know that X=15. We have to find a Y value so that the fraction with an x of 15 simplified is 8/3. To do this you would write 8/3 and y/15. Now cross multiply to get 3y=120. Divide by 3 to get y=40. View my attachment for the work!
Answer:
so its 50+.25(x) = y or 140. x is amount of minutes y is total
Answer:
yes
Step-by-step explanation:
The line intersects each parabola in one point, so is tangent to both.
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For the first parabola, the point of intersection is ...
y^2 = 4(-y-1)
y^2 +4y +4 = 0
(y+2)^2 = 0
y = -2 . . . . . . . . one solution only
x = -(-2)-1 = 1
The point of intersection is (1, -2).
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For the second parabola, the equation is the same, but with x and y interchanged:
x^2 = 4(-x-1)
(x +2)^2 = 0
x = -2, y = 1 . . . . . one point of intersection only
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If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.
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Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.