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matrenka [14]
3 years ago
10

Tell whether the lines for each pair of equations are parallel, perpendicular, or neither

Mathematics
1 answer:
stiks02 [169]3 years ago
4 0

Answer:

C)  Neither

Step-by-step explanation:

y = -1/4x + 8

-2x + 8y = 4

To compare the slopes  we need to convert the second equation to the same form as the first ( that is slope/intercept form).

-2x + 8y = 4

8y = 2x + 4

Divide through by 8:-

y = 1/4 x + 1/2

The slope of the first line is -1/4 and its  1/4 for the second line.

If the slopes were the same the lines would be parallel and if they were perpendicular  their product would be = -1.

The product is 1/4 * -1/4 = -1/16

So they are neither parallel nor perpendicular.

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The radius of a quarter circle is 3 what is the quarter circle's perimeter?
slava [35]

Answer:

21.42

Step-by-step explanation:

Imagine it as a full circle. 3 x 4 will give you the diameter of a full circle, which is 12. circumference = pi x diameter. so 12 pi (we can work that out later). now for the quater circle, divide it by 4, so 3 pi. Then you have the 2 extra edges which are both 2r, so 6 each. finally, you have 3 pi plus 12 which = 21.42

6 0
1 year ago
What is 50% of 33 show work please​
olganol [36]

Answer:

16.5

Step-by-step explanation:

50 x 33

-------------

100

= 16.5

8 0
2 years ago
One more than three-eighths of a number is eleven. Just write out equation do not solve please
alexdok [17]

Answer:

Let's just call the number x. 3/8ths of x and 1 more equals 11.

As an equation, this basically means

\frac{3}{8} x+1=11

4 0
3 years ago
Pleaseee help. Oh yeah this has to be 20 characters soooo
valina [46]

It is more than one I think because it becomes -1 over -12 WITHOUT the -power and then add the power and it becomes positive. Make the power positive and it will become negative making it less than 1.

6 0
3 years ago
Read 2 more answers
Find an equation of the tangent line to the curve 2(x2+y2)2=25(x2−y2) (a lemniscate) at the point (−3,1). An equation of the tan
valina [46]

2(x^2+y^2)^2=25(x^2-y^2)

Let y=y(x), so that differentiating both sides wrt x gives

4(x^2+y^2)\left(2x+2y\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(2x-2y\dfrac{\mathrm dy}{\mathrm dx}\right)

If x=-3 and y=1, the above reduces to

40\left(-6+2\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(-6-2\dfrac{\mathrm dy}{\mathrm dx}\right)\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac9{13}

This is the slope of the tangent line, which has equation

y-1=\dfrac9{13}(x+3)\implies\boxed{y=\dfrac{9x+40}{13}}

7 0
2 years ago
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