<span> £536,521.
</span>6,521.
6,000
the thousands place
Answer:
I think C 45 blocks are the right answer.
Answer:
1) B = 66.5° c = 10.9
Step-by-step explanation:
I will do question one as an example. In general, for these questions you want to use the appropriate trigonometric ratios to solve for the variables and/or apply logic using rules regarding triangles. See attached image for all solving steps.
For side c, we can use Cosine of angle A for a ratio between 10 and c. When we write out the equation, we can solve for side c. So when we write it out, we get the equation:
cos23.5 = ¹⁰⁄c
c = ¹⁰⁄cos₂₃.₅
c = 10.9044 (make sure to round to the nearest tenth, which is one decimal place)
For angle B, since they have given two angles, you can solve for B since all angles of a triangle add up to 180 degrees.
So b = 180 - (90 + 23.5) = 180 - 113.5
b = 66.5
- It is also possible to solve this using sine of angle B and solve it from there, but applying the theory this way is much simpler. (this is on the image if you're curious about it)
I hope this helps you with the other 3 questions.
Answer:
<h3>
Lucas and Zoe will both next visit their mother on 7th of May at the same time.</h3>
Step-by-step explanation:
<em>First</em><em>,</em><em> </em><em>find </em><em>the </em><em>first</em><em> </em><em>multiple </em><em>that </em><em>can </em><em>be </em><em>divisible </em><em>by </em><em>5</em><em> </em><em>and </em><em>8</em><em>:</em>
<em>4</em><em>0</em><em> </em><em>is </em><em>divisible </em><em>by </em><em>5 </em><em>and </em><em>8</em><em>.</em>
<em>Second</em><em>,</em><em> </em><em>find </em><em>the </em><em>days </em><em>in </em><em>April </em><em>and </em><em>subtract </em><em>7</em><em> </em><em>from </em><em>the </em><em>days:</em>
<em>April</em><em>=</em><em>3</em><em>0</em><em> </em><em>days</em><em>.</em>
<em>3</em><em>0</em><em>-</em><em>7</em><em>=</em><em> </em><em>2</em><em>3</em><em> </em><em>days</em><em>.</em>
<em>Finally,</em><em> </em><em>subtract </em><em>2</em><em>3</em><em> </em><em>from </em><em>4</em><em>0</em><em> </em><em>and </em><em>find </em><em>it </em><em>in </em><em>the </em><em>next </em><em>month</em><em>,</em><em> </em><em>which </em><em>is </em><em>May</em><em>:</em>
<em>4</em><em>0</em><em>-</em><em>2</em><em>3</em><em>=</em><em>1</em><em>7</em>
<em>Therefore</em><em>,</em><em> </em><em>they </em><em>will </em><em>both </em><em>next </em><em>visit </em><em>their </em><em>mother </em><em>on </em><em>the </em><em>7</em><em>t</em><em>h</em><em> </em><em>of </em><em>May.</em>
Let M = {Λ,abb} and L = {bba,ab, a}, what is ML ? ML ={bba, abbbba,abbab,abbba, ab,a} ML ={bba, abbbba,abbab,abba, ab,a} ML ={bb
Xelga [282]
Answer:
ML = {bba, ab, a, bbaabb, ababb, aabb}
Step-by-step explanation:
By application of Union of a set.
M = {bba,ab, a}
L = {Λ,abb}
ML = {bba, ab, a, bbaabb, ababb, aabb}
