In the diagram below, transversal t intersects parallel lines I and m.
1 answer:
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<u><em>Answer:</em></u>
D. f⁻¹(x) = 2x + 10
<u><em>Explanation:</em></u>
<u>The given is:</u>
<u>To find the inverse, follow the below procedures:</u>
<u>1- Convert f(x) to y. </u>
<u>This will give:</u>
<u>2- Swap the x and y variables.</u>
<u>This will give:</u>
<u>3- Solve for y.</u>
<u>This will give:</u>
<u>4- Replace the y in the above step with f⁻¹(x).</u>
<u>This will give us:</u>
f⁻¹(x) = 2x + 10
Hope this helps :)
The bridge attached is drawn according to given dimensions, and it doesn't look right. Please double check the given dimensions.
Calculations:
Horizontal part of bottom chord below the 70 degree triangle
= 15.1*cos(70) = 5.16 (which is a major prt of the 6.3 units.
Height of vertical pieces DF and EH
= 15.1*sin(70) = 14.19
Note that structurally, DF and EH do not help in reducing stress on the bridge, since they are perpendicular to the bottom chord.
Therefore
angle B = atan(14.19/(6.3-5.16))=85.41 degrees
I believe the whole geometry does not look right, esthetically, and structurally, since the compression members are much longer than the tension members in the middle. (The vertical members carry no force.)
If you can review the input data, or post a new question, I will be glad to help.
Calculations:
Horizontal part of bottom chord below the 70 degree triangle
= 15.1*cos(70) = 5.16 (which is a major prt of the 6.3 units.
Height of vertical pieces DF and EH
= 15.1*sin(70) = 14.19
Note that structurally, DF and EH do not help in reducing stress on the bridge, since they are perpendicular to the bottom chord.
Therefore
angle B = atan(14.19/(6.3-5.16))=85.41 degrees
I believe the whole geometry does not look right, esthetically, and structurally, since the compression members are much longer than the tension members in the middle. (The vertical members carry no force.)
If you can review the input data, or post a new question, I will be glad to help.