Answer:
6th period.
Step-by-step explanation:
Let the one type of the bread be bread A
The second type of the bread be bread B
Let the flour be 'f' and the butter be 'b'
We need 150f + 50b for bread A and 75f + 75b for bread B
We can compare the amount of flour and bread needed for each bread and write them as ratio
FLOUR
Bread A : Bread B
150 : 75
2 : 1
We have a total of 2250gr of flour, and this amount is to be divided into the ratio of 2 parts : 1 part. There is a total of 3 parts.
2250 ÷ 3 = 750 gr for one part then multiply back into the ratio to get
Bread A : Bread B = (2×750) : (1×750) = 1500 : 750
BUTTER
Bread A : Bread B = 50 : 75 = 2 : 3
The amount of butter available, 1250 gr is to be divided into 2 parts : 3 parts.
There are 5 parts in total
1250 ÷ 5 = 250 gr for one part, then multiply this back into the ratio
Bread A: Bread B = (2×250) : (3×250) = 500 : 750
Hence, for bread A we need 1500 gr of flour and 500 gr of butter, and for bread B, we need 750 gr of flour and 750 gr of butter.
The answer is A. Slope is rise/run. According to the graph, you can trace out and see that it rises 1 and goes to the right 4 times. This means that it has a positive slope of 1/4. And the line crosses the y intercept at y = 3
So answer choice A is correct.
I did this manually on some notepaper but my scanner is not working.
The answer is 72 cm^2
Answer:
18
Step-by-step explanation:
<em>Convert</em><em> </em><em>the</em><em> </em><em>mixed</em><em> </em><em>number</em><em> </em><em>to</em><em> </em><em>an</em><em> </em><em>improper</em><em> </em><em>fraction</em><em> </em><em>9</em><em>/</em><em>2</em><em>÷</em><em>1</em><em>/</em><em>4</em>
<em>Reduce</em><em> </em><em>the</em><em> </em><em>numbers</em><em> </em><em>with</em><em> </em><em>the</em><em> </em><em>greatest</em><em> </em><em>common</em><em> </em><em>factor</em><em> </em><em>2</em><em> </em>
<em>and</em><em> </em><em>then</em><em> </em><em>multiply</em><em> </em><em>the</em><em> </em><em>numbers</em><em> </em><em>9</em><em>×</em><em>2</em><em>=</em><em>18</em><em> </em><em>✅</em>