Answer:
<em>I </em><em>think</em><em> the</em><em> two</em><em> </em><em>digit </em><em>number</em><em> </em><em>formed </em><em>would </em><em>be </em>
<em>xy</em>
<em>seeing</em><em> </em><em>that </em><em>x </em><em>is </em><em>on </em><em>the </em><em>tens </em><em>place </em><em>which </em><em>is </em><em>the </em><em>second</em><em> </em><em>last </em><em>place </em><em>and </em><em>y </em><em>is </em><em>on </em><em>the </em><em>ones </em><em>place </em><em>meaning</em><em> </em><em>it </em><em>has </em><em>to </em><em>be </em><em>the </em><em>last </em><em>value.</em>
<em>I</em><em> hope</em><em> this</em><em> helps</em>
Answer:
What's the question.?????????
Answer:
For the first
Since the angles in a triangle sum up to 180.
We already have angleA 41° and angleC 90°
To get the remaining angle B 180-(41+90)=49°
Now
We apply sine rule to solve
Divide each term you pick by the Sine of the angle directly opposite it.
So We have
4/Sin41° = X/Sin49
Making X subject..
X= 4Sin49/Sin41
X=4.601
To the nearest tenth
X= 4.6.
For the second question
Applying the same concept
Angle B = 180-(90 + 47) = 43°
Applying Sine Rule
X/Sin43 = 3/Sin90
Making x subject
X= 3Sin43/Sin90
X=2.04
To the nearest Tenth
X=2.0
