C= two step equations require all number to be whole number in order to solve.
Here is an example that disproves that:
.5x + 8= 18
<u> -8 -8</u>
<u>.5</u>x = <u>10
</u>.5 .5
x = 20 (.5 is certainly not a whole number but I solved it anyway)
Answer is : 1.2 geometric series
<h2>
Answer:</h2>
cos 28°cos 62°– sin 28°sin 62° = 0
<h2>
Step-by-step explanation:</h2>
From one of the trigonometric identities stated as follows;
<em>cos(A+B) = cosAcosB - sinAsinB -----------------(i)</em>
We can apply such identity to solve the given expression.
<em>Given:</em>
cos 28°cos 62°– sin 28°sin 62°
<em>Comparing the given expression with the right hand side of equation (i), we see that;</em>
A = 28°
B = 62°
<em>∴ Substitute these values into equation (i) to have;</em>
<em>⇒ cos(28°+62°) = cos28°cos62° - sin28°sin62°</em>
<em />
<em>Solve the left hand side.</em>
<em>⇒ cos(90°) = cos28°cos62° - sin28°sin62°</em>
⇒ 0 = <em>cos28°cos62° - sin28°sin62° (since cos 90° = 0)</em>
<em />
<em>Therefore, </em>
<em>cos28°cos62° - sin28°sin62° = 0</em>
<em />
<em />
Finding the upper and lower bounds for a definite integral without an equation is pretty hard because how can we find the upper and lower bounds of definite integral if there is no equation given. But I will teach you how to find the lower and upper bounds of a definite integral, when the equation is like this
So, i integrate this,
I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4
