What we need to do is divide the total prize by the number of units, that is by 9:
we know this because 9 of those pieces, or 9p is equal to 54.90
so the right equation is <span>B. 9p= 54.90
(another possibility would be p=54.90/9)</span>
Answer:
19
Step-by-step explanation:
<em>this </em><em>question</em><em> </em><em>means </em><em>you </em><em>have </em><em>to </em><em>place </em><em>the </em><em>value </em><em>of </em><em>-</em><em>7</em><em> </em><em>were </em><em>there's</em><em> </em><em>x </em><em>in </em><em>the </em><em>function</em><em>.</em>
<em>f(</em><em>-</em><em>7</em><em>)</em><em>=</em><em>-</em><em>3</em><em>(</em><em>-</em><em>7</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>2</em><em>1</em><em>-</em><em>2</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>1</em><em>9</em>
<em>I </em><em>hope</em><em> this</em><em> helps</em>
Complete question :
A, B, C and D are four towns. B is 30 kilometres due east of A. C is due 30 kilometres north of A. D is due 45 kilometres due south of A. Work out the bearing of B from C.
Answer:
44°
Step-by-step explanation:
From the diagram sketched, we have a right angled d triangle, thus we can apply Pythagoras ;
Using the relation :
Tanθ = opposite / Adjacent
Tan θ = 30 / 30
Tanθ = 1
θ = tan^-1(1)
θ = 45°
Hence, bearing of B from C is 45°
-3x-8=10
First add 8 to both sides
-3x=18
Then divide both sides by -3
x=-6
it is 3 numbers all together and they are 25, 36 and 49.