Please help me solve this it’s I center

Think about it this way:

triangle KDI is supplementary to EHK. SO if we find the angles of KDI we can do the same for EHK. Additionally, the m<H in the triangle HKF is 90 degrees and since it is supplementary to EHK then the measure of h in EHK is also 90 degrees. So m<h is 90. All three angles will equal to 180 since the interior angles of any triangle equal 180°

Now, lets find the measures of KDI so we can solve for the measure of K in the triangle EHK

D in KDI is 19 degrees and the square on I in KDI indicates its a 90 degree angle. Those add up to 109 degreess meaning measure of k must equal 71 degrees (180=109=71) since the interior angles of any triangle equal 180. Therefor in the triangle EHK, h is 90, and k is 71 degrees by vertical angles theorem. 90 and 71 add up to 161 degrees meaning angle E is going to be equal to 19 (180-161=19). this, therefore means 17x-7=19

solve:

17x-7=19

17x=26

x=26/17

x=1 9/17

You can write it as a decimal, improper fraction or mixed number.

Step-by-step explanation:

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

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2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

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∪

[

0

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∞

)

{

y

|

y

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step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.