Answer:
The equation is given below as

Step 1:
We will work on the left-hand side, we will have

By substituting the identity above, we will have

Here, we will make use of the quotient identity
Step 2:
By writings an expression, we will have

Here, we will use the definition of subtraction

Step 3:
We will apply the double number identity given below

By applying this, we will have

Here, we will use the double number identity
x ≤ 12
Solution:
Let us take the number be x.
A number : x
A number divided by three : 
A number divided by three less two : 
A number divided by three less two is at most two : 
Now, simplify the inequality,

Add 2 on both sides of the inequality,


Multiply by 3 on both sides of the equation,


Hence x ≤ 12.
Answer:
The discrimant of this equation is 144.
Step-by-step explanation:
First you have to move all the variables to one side to make the equation/expression into 0 by substracting 8x² to both sides :
9x² + 14x + 13 = 8x²
9x² + 14x + 13 - 8x² = 8x² - 8x²
x² + 14x + 13 = 0
It is given that the formula of discriminant is, D = b² - 4ac where a&b&c represent the number of the equation, ax²+bx+c = 0 :
x² + 14x + 13 = 0
D = b² - 4ac
= 14² - 4(1)(13)
= 196 - 52
= 144
Answer:
0a23,02A3
Step-by-step explanation:
I think that this is the right answer Im sorry if it is wrong
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.