Answer:
In first box: top left __7, in top right __4, on left top __9, left bottom __6.
In second box: top left __7, in top right __3, on left top __3, left bottom __5.
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
Given
+ ![\left[\begin{array}{ccc}3&1\\-1&2\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%261%5C%5C-1%262%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Add corresponding elements to obtain the sum, that is
= ![\left[\begin{array}{ccc}-2+3&3+1\\2-1&4+2\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%2B3%263%2B1%5C%5C2-1%264%2B2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=
→ C
Answer:
y=kx
subst y=10 and x=20 into the above
<em>1</em><em>0</em><em>=</em><em>k20</em>
<em>k</em><em>=</em><em>1</em><em>0</em><em>/</em><em>2</em><em>0</em>
<em>k</em><em>=</em><em>1</em><em>/</em><em>2</em>
<em>therefore</em><em> </em><em>relationship</em><em>:</em><em> </em><em>y</em><em>=</em><em>1</em><em>/</em><em>2</em><em>x</em>
<em>subst</em><em> </em><em>x</em><em>=</em><em>1</em><em>5</em><em> </em><em>into</em><em> </em><em>the</em><em> </em><em>relationship</em>
<em> </em><em>y</em><em>=</em><em>1</em><em>/</em><em>2</em><em>(</em><em>1</em><em>5</em><em>)</em>
<em>y</em><em>=</em><em>7</em><em>,</em><em>5</em>
Step by step explanation:
- Step 1: when they say y varies directly with x they mean<em> y is proportional to x</em>
- step 2: so y=kx where <em>k is the constant</em>
- step 3: is to substitute <em>y=10</em> and <em>x=20</em> into the above equation y=kx
- step 4: you will end up with <em>10=k20</em> then divide both sides by 20 so that <em>k becomes the subject of the formula </em>
- step 5: your answer from the above will be <em>k=10/20 </em>so the relationship is <em>y is directly proportional to 1/2 x </em>what you did here is that you substituted k for 1/2 in the equation in step 3
- step 6: is to finally substitute x=15 into the equation <em>y=1/2x</em> to finally get your answer <em>y</em><em>=</em><em>7</em><em>,</em><em>5</em><em>.</em>