The 20th term in the following arithmetic sequence1/2, 1, 3/2, 2... is 10 because;
the number are divided by 2
and the 20th number would be 20/2 which is equal to 10
I did this test b4, yours is answer #number 12
Convert things to their basic forms.
<span>Remember a few identities </span>
<span>sin^2 + cos^2 = 1 so </span>
<span>sin^2 = 1 - cos^2 and </span>
<span>cos^2 = 1 - sin^2 </span>
<span>I'm going to skip typing the theta symbol, just to make things faster. Just assume it is there and fill it in as you work the problems. </span>
<span>Follow along to see how each problem was worked out. You'll catch on to the general technique. </span>
<span>====== </span>
<span>1. sec θ sin θ </span>
<span>1/cos * sin = sin/cos = tan </span>
<span>2. cos θ tan θ </span>
<span>cos * sin/cos = sin </span>
<span>3. tan^2 θ- sec^2 θ </span>
<span>sin^2 / cos^2 - 1/cos^2 </span>
<span>(sin^2 - 1)/cos^2 </span>
<span>-(1-sin^2)/cos^2 </span>
<span>-cos^2/cos^2 </span>
<span>-1 </span>
<span>4. 1- cos^2θ </span>
<span>sin^2 </span>
<span>5. (1-cosθ)(1+cosθ) </span>
<span>Remember (a+b)(a--b) = a^2 - b^2 </span>
<span>1-cos^2 = sin^2 </span>
<span>6. (secx-1) (secx+1) </span>
<span>sec^2 -1 </span>
<span>1/cos^2 - 1 </span>
<span>1/cos^2 - cos^2/cos^2 </span>
<span>(1-cos^2)/cos^2 </span>
<span>sin^2/cos62 </span>
<span>tan^2 </span>
<span>7. (1/sin^2A)-(1/tan^2A) </span>
<span>1/sin^2 - 1/(sin^2/cos^2) </span>
<span>1/sin^2 - cos^2/sin^2 </span>
<span>(1-cos^2)/sin^2 </span>
<span>sin^2/sin^2 </span>
<span>1 </span>
<span>8. 1- (sin^2θ/tan^2θ) </span>
<span>1-sin^2/(sin^2/cos^2) </span>
<span>1 - sin^2*cos^2/sin^2 </span>
<span>1-cos^2 </span>
<span>sin^2 </span>
<span>9. (1/cos^2θ)-(1/cot^2θ) </span>
<span>1/cos^2 - 1/(cos^2/sin^2) </span>
<span>1/cos^2 - sin^2/cos^2 </span>
<span>(1-sin^2)/cos^2 </span>
<span>cos^2/cos^2 </span>
<span>1 </span>
<span>10. cosθ (secθ-cosθ) </span>
<span>cos *(1/cos - cos) </span>
<span>1-cos^2 </span>
<span>sin^2 </span>
<span>11. cos^2A (sec^2A-1) </span>
<span>cos^2 * (1/cos^2 - 1) </span>
<span>1 - cos^2 </span>
<span>sin^2 </span>
<span>12. (1-cosx)(1+secx)(cosx) </span>
<span>(1-cos)(1+1/cos)cos </span>
<span>(1-cos)(cos + 1) </span>
<span>-(cos-1)(cos+1) </span>
<span>-(cos^2 - 1) </span>
<span>-(-sin^2) </span>
<span>sin^2 </span>
<span>13. (sinxcosx)/(1-cos^2x) </span>
<span>sin*cos/sin^2 </span>
<span>cos/sin </span>
<span>cot </span>
<span>14. (tan^2θ/secθ+1) +1 </span>
<span>(sin^2/cos^2)/(1/cos) + 2 </span>
<span>sin^2/cos + 2 </span>
<span>sin*tan + 2 </span>
P11,p7 ?
I dont really understand what that is so.
Answer:
m∠JLK° = 62°
Step-by-step explanation:
∠JLK = ∠KJL
SO
56°+2(∠JLK)° = 180°
2(∠JLK)° = 180-56
2(∠JLK)° = 124
∠JLK° = 62°
so m∠JLK° = 62°
Answer:
A. The situation is discrete B. i. { x : 0 ≤ x ≤ 6; x ∈ Z} ii. { C = 5x : 0 ≤ C ≤ 30; C ∈ Z}
Step-by-step explanation:
A. The situation is discrete since we have integral values for the amount paid per mile walked. The amount per mile is $5 and is only paid if a complete mile is walked. So, it is a discrete situation.
B. i. Since 0 miles represents 0 distance and the student walks a maximum of 6 miles, let x represent the distance walked. So the domain is 0 ≤ x ≤ 6 where x ∈ Z where Z represent integers.
{ x : 0 ≤ x ≤ 6; x ∈ Z}
ii. Since at 0 miles the amount earned is 0 miles × $5 per mile = $ 0 and at the maximum distance of 6 miles, the amount earned is 6 miles × $5 per mile = $ 30, let C represent the amount donated in dollars. So the range is 0 ≤ C ≤ $ 30 where C = 5x.
{ C = 5x : 0 ≤ C ≤ 30; C ∈ Z}
