Answer:
=0.4245 (4dp)
Step-by-step explanation:
tan(23)=0.4244748162
=0.4245 (4dp)
I hope this helps!
Answer:
(3x-1)(2x+1)
Step-by-step explanation:
1-6x^2-x=0
-6x^2-x+1=0
6x^2+x-1=0
6x^2+(3-2)x-1=0
6x^2+3x-2x-1=0
3x(2x+1)-1(2x+1)=0
(3x-1)(2x+1)=0
So the factor is (3x-1)(2x+1)
Step 1
<u>Find the value of TS</u>
we know that
if PQ is parallel to RS. then triangles TRS and TPQ are similar
so
![\frac{TR}{TP} =\frac{TS}{QT}](https://tex.z-dn.net/?f=%5Cfrac%7BTR%7D%7BTP%7D%20%3D%5Cfrac%7BTS%7D%7BQT%7D)
solve for TS
![TS =\frac{TR*QT}{TP}](https://tex.z-dn.net/?f=TS%20%3D%5Cfrac%7BTR%2AQT%7D%7BTP%7D)
we have
![RP=2\ cm\\TP=18\ cm\\QT=27\ cm](https://tex.z-dn.net/?f=RP%3D2%5C%20cm%5C%5CTP%3D18%5C%20cm%5C%5CQT%3D27%5C%20cm)
![TR=TP+RP\\TR=18+2=20\ cm](https://tex.z-dn.net/?f=TR%3DTP%2BRP%5C%5CTR%3D18%2B2%3D20%5C%20cm)
substitute
![TS =30\ cm](https://tex.z-dn.net/?f=TS%20%3D30%5C%20cm)
Step 2
<u>Find the value of SQ</u>
we know that
![SQ=TS-QT](https://tex.z-dn.net/?f=SQ%3DTS-QT)
we have
![TS =30\ cm](https://tex.z-dn.net/?f=TS%20%3D30%5C%20cm)
![QT=27\ cm](https://tex.z-dn.net/?f=QT%3D27%5C%20cm)
substitute
![SQ=30\ cm-27\ cm=3\ cm](https://tex.z-dn.net/?f=SQ%3D30%5C%20cm-27%5C%20cm%3D3%5C%20cm)
therefore
<u>the answer is</u>
the value of SQ is ![3\ cm](https://tex.z-dn.net/?f=3%5C%20cm)
Step-by-step explanation:
s1 = 3
s2 = 3×2 = 6
s3 = 6×2 = 12
s4 = 12×2 = 24
s5 = 24×2 = 48
sn = sn-1 × 2 = s1 × 2^(n-1)
s17 = 3 × 2¹⁶ = 3 × 65,536 = 196,608
A cross section is a shape formed by cutting an object using a plane. For this case, a cone is sliced parallel to the slant height so the cross section formed would be a plane with a U-like shape. From the image attached, the correct answer is the second option.