Step-by-step explanation:
1. Use the Pythagorean theorem to solve for side KM:
- 16^2+KM^2=34^2
- 256+KM^2=1156
- KM^2=900
- KM=30
2. Cosine is adjacent over hypotenuse, so cosine of M would be KM/34, or 30/34
3. Tangent is opposite over adjacent, so tangent of L will be KM/16, or 30/16
4. Sine is opposite over hypotenuse, so sine of M will be 16/34
5. KM=30, solved for in step 1.
hope this helps!!
Answer:
1) y=⅚x -2⅓
2) y=8/3x -5
Step-by-step explanation:
<u>Point-slope form:</u>
y=mx+c, where m is the gradient and c is the y-intercept.
Parallel lines have the same gradient.
Gradient of given line= 
Thus, m=⅚
Susbt. m=⅚ into the equation,
y= ⅚x +c
Since the line passes through the point (4, 1), (4, 1) must satisfy the equation. Thus, substitute (4, 1) into the equation to find c.
When x=4, y=1,
1= ⅚(4) +c

Thus the equation of the line is
.
The gradients of perpendicular lines= -1.
Gradient of given line= -⅜
-⅜(gradient of line)= -1
gradient of line
= -1 ÷ (-⅜)
= -1 ×(-8/3)
= 

When x=3, y=3,

Thus the equation of the line is
.
Answer:
61,940
Step-by-step explanation:
For a recursive sequence of reasonable length, it is convenient to use a suitable calculator for figuring the terms of it. Since each term not only depends on previous terms, but also depends on the term number, it works well to use a spreadsheet for doing the calculations. The formula is easily entered and replicated for as many terms as may be required.
__
The result of executing the given algorithm is shown in the attachment. (We have assumed that g_1 means g[-1], and that g_2 means g[-2]. These are the starting values required to compute g[0] when k=0.
That calculation looks like ...
g[0] = (0 -1)×g[-1] +g[-2} = (-1)(9) +5 = -4
The attachment shows the last term (for k=8) is 61,940.
Attach a photo so we can see the problem
<h2>
<em>Answer:</em></h2><h2>
<em>9</em><em> </em><em>pi </em><em>m^</em><em>2</em></h2>
<em>Solution</em><em>,</em>
<em>Diameter</em><em>=</em><em>6</em><em> </em><em>m</em>
<em>Radius=</em><em> </em><em>6</em><em>/</em><em>2</em><em>=</em><em>3</em><em>m</em>
<em>Area </em><em>of </em><em>circle=</em><em> </em><em>pi </em><em>r </em><em>^</em><em>2</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>pi </em><em>*</em><em>(</em><em>3</em><em>)</em><em>^</em><em>2</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>pi*</em><em>9</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>9</em><em> </em><em>pi </em><em>metre^</em><em>2</em>
<em>Hope </em><em>it</em><em> helps</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>