Hey there! :D
So you're wondering if you can make a new ten in this problem, right? Well, lets find out! We can simply do<span> 59 + 17 = 59 + 1 + 16 = 60 + 16 = 76. When we did 59 + 1, that is a new ten then we added to 16. So yes, we did make a new ten in this problem. </span>
<span>Hope it helps! ;)</span>
Answer:
aby/2 = c
Step-by-step explanation:
1) ab * y = ab/2c
2) aby/2 = 2c/2
Final answer: aby/2 = c
Answer:
h = 10sin(π15t)+35
Step-by-step explanation:
The height of the blade as a function f time can be written in the following way:
h = Asin(xt) + B, where:
B represets the initial height of the blade above the ground.
A represents the amplitud of length of the blade.
x represents the period.
The initial height is 35 ft, therefore, B = 35ft.
The amplotud of lenth of the blade is 10ft, therefore A = 10.
The period is two rotations every minute, therefore the period should be 60/4 = 15. Then x = 15π
Finally the equation that can be used to model h is:
h = 10sin(π15t)+35
Answer:
(3, 0).
Step-by-step explanation:
dentifying the vertices of the feasible region. Graphing is often a good way to do it, or you can solve the equations pairwise to identify the x- and y-values that are at the limits of the region.
In the attached graph, the solution spaces of the last two constraints are shown in red and blue, and their overlap is shown in purple. Hence the vertices of the feasible region are the vertices of the purple area: (0, 0), (0, 1), (1.5, 1.5), and (3, 0).
The signs of the variables in the contraint function (+ for x, - for y) tell you that to maximize C, you want to make y as small as possible, while making x as large as possible at the same time.
Hence, The Answer is ( 3, 0)
Answer:
<h2>
<em>y</em><em>=</em><em>-</em><em>4</em></h2>
<em>Sol</em><em>ution</em><em>,</em>
<em>X=</em><em>4</em>
<em>Now</em><em>,</em>
<em>
</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em>