(5x + 12) 48° solve for x.
1 answer:
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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.
Answer:
Step-by-step explanation:
Answer:
a) The coordinates are (0.431, -2.646) and (3.568,-7.354)
b) The tangent lines are
l₁ = (x-0.431)*2/3 - 2.646
l₂ = (x-3.568)2/3 - 7.354
Step-by-step explanation:
First, lets complete squares, by taking for each cordinate the square of a linear expression
0= x²−4x+y²+10y+13 = (x-2)²+ 4 + (y+5)²-25+13 = (x-2)²+(y+5)² - 8
Hence (x-2)² + (y+5)² = 8
Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.
(y+5)² = 8 - (x-2)² = -x² + 4x + 4
Thus
Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.
We may just find the values of x which the derivate of f is either 2/3 or -2/3.
The quadratic has roots
one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.
Also g(3.568) = - 7.354 and f(0.431) = -2.646
This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are
l₁ = (x-0.431)*2/3 - 2.646
l₂ = (x-3.568)2/3 - 7.354