<span>g(2+h)=5+m
g=(5+m)/(2+h)</span>
Step-by-step explanation:
Let's take the RHS,
we've,
<h3>(<u>Cosa</u><u>/</u><u>2</u><u> </u><u>-</u><u> </u><u>sina</u><u>/</u><u>2</u><u>)</u></h3><h3>(<u>Cosa/</u><u>2</u><u> </u><u>+</u><u> </u><u>sina</u><u>/</u><u>2</u><u>)</u></h3>
Let's Rationalise the Denominator.
we get,
<h3>(<u>Cosa/</u><u>2</u><u> </u><u>-</u><u> </u><u>sina</u><u>/</u><u>2</u><u>)</u><u>^</u><u>2</u></h3><h3><u>(</u><u>cosa</u><u>/</u><u>2</u><u>)</u><u>^</u><u>2</u><u> </u><u>-</u><u> </u><u>(</u><u>sina</u><u>/</u><u>2</u><u>)</u><u>^</u><u>2</u></h3>
The numerator is in form of (a-b)^2 and the denominator is in form of a^2-b^2. Now,
By formula,
<h3>(<u>Cosa/</u><u>2</u><u>)</u><u>^</u><u>2</u><u> </u><u>-2cosa</u><u>/</u><u>2</u><u>.</u><u>s</u><u>i</u><u>n</u><u>a</u><u>/</u><u>2</u><u> </u><u>+</u><u> </u><u>(</u><u>sina</u><u>/</u><u>2</u><u>)</u><u>^</u><u>2</u> </h3><h3> cosa</h3>
Here I substituted Cosa in place of (Cosa/2)^2 - (sina/2)^2 because it's the formula of cosa in sub multiple angle form.
<h3>In the numerator, </h3>
(sina/2)^2 + (Cosa/2)^2 =1.........( by formula)
so we have,
<h3><u>1</u><u> </u><u>-</u><u> </u><u>2</u><u>s</u><u>i</u><u>n</u><u>a</u><u>/</u><u>2</u><u>.</u><u>c</u><u>o</u><u>s</u><u>a</u><u>/</u><u>2</u></h3><h3>Cosa</h3>
<h3 /><h3 /><h3><u>1</u><u> </u><u>-</u><u> </u><u>sina</u> {because 2sina/2.cosa/2=sina)</h3><h3>Cosa</h3>
LHS proved.
Thank You.
It is 70 hope this helps :)
Answer:
Yes because
Step-by-step explanation:
Is 3z-1 in standard form?
Yes becausestandard form of a polynomial, the value in the power of the variable such as z in the equation reduces from left to right. Therefore, 3z - 1 is in standard form since the z term comes to the left of the constant term