Jenny has saved $600 from her summer job. During the school year, she withdraws $20 from her account every week for her daily ex
2 answers:
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Answer:
<u>Volume</u>
For the rectangle, h = 3cm, l = 8cm, w = 6cm
V = length x width x height
V = 8cm x 6cm x 3cm
V = 144cm^3
For the semi circle, we need to find the radius. The radius is width/2, so 6cm/2 = 3cm. r = 3cm, = 3.14
V = radius^2 x height x
V = 3cm^2 x 3cm x 3.14
V = 84.8 cm^3/2 (because the cylinder needs to be divided to form a semi-circle)
V= 42.4cm^3 (there are two cylinders though so we will multiply this by 2 in the total volume)
Total volume:
V = 144cm^3 + 42.4cm^3(2)
V = 186.4cm^3
<u>Surface Area</u>
Rectangular prism:
A = 2[w(l) + h(l) + h(w)]
A = 2[6cm(8cm) + 3cm(8cm) + 3cm(6cm)]
A = 180cm^2
But there are two sides that are covered by the semi-circular prisms, so we will have to calculate those sides and remove them.
A = l x w
A = 6cm x 3cm
A = 18cm^2(2) (2 being the two faces)
A = 36cm^2
A = 180cm^2 - 36cm^2
A = 144cm^2 (the area of the rectangle)
Semi-circular prism:
A = 2rh + 2
r^2
Earlier, we found out that the radius of the circle is 3cm, so we will plug that in.
A = 2(3.14)(3cm)(3cm) + 2(3.14)(3cm)^2
A = 113.09cm^2
Total surface area:
A = 144cm^2 + 133.09cm^2
A = 277.09cm^2
Therefore the total volume of the prism is 186.4cm^3 and the total surface area is 277.09cm^2.
Answer:
1)
Minimum is a 4th Degree
2)
Positive; Even
3)
Step-by-step explanation:
Part 1)
The minimum degree of our function will be 4.
Looking at the graph, we know that the graph crosses the x-axis at -4 and -1. Since it <em>crosses through</em> the x-axis at these two points, these two factors must have an odd multiplicity.
So, it can be anything 1, 3, 5, 7, etc.
However, we will choose the lowest one, 1.
Next, we know that the graph <em>bounces off</em> at 3.
So, it must have an even multiplicity. In other words, 2, 4, 6, 8, etc.
We choose the lowest one, 2.
Therefore, the minimum degree of our function will be 1+1+2 or 4.
Part 2)
The degree of our polynomial is (and will always be) even. Therefore, both ends of the graph will go in the same direction.
Recall the simplest even polynomial, the parent quadratic function. When the leading coefficient is positive, both of the ends go straight up.
This applies to all polynomials with even degrees.
Therefore, since the arms of the graph is going straight up towards positive infinity, the leading coefficient of our graph must be positive.
Part 3)
This is similar to Part 1.
We can see that the graph touches the x-axis at -4, -3, and 1. So, the zeros of the function is:
We know that it<em> passes through</em> . So, these two factors must have an odd multiplicity.
However, since the graph <em>bounces off</em> , this factor must have an even multiplicity.
And we're done!