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 + 2r^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:
answer is 10
Step-by-step explanation:
Because it is 38 feet right? and the building is 25 feet. 38-25=13. round it to nearest tenth, you get 10.
<span>Answer:
Its too long to write here, so I will just state what I did.
I let P=(2ap,ap^2) and Q=(2aq,aq^2)
But x-coordinates of P and Q differ by (2a)
So P=(2ap,ap^2) BUT Q=(2ap - 2a, aq^2)
So Q=(2a(p-1), aq^2)
which means, 2aq = 2a(p-1)
therefore, q=p-1
then I subbed that value of q in aq^2
so Q=(2a(p-1), a(p-1)^2)
and P=(2ap,ap^2)
Using these two values, I found the midpoint which was:
M=( a(2p-1), [a(2p^2 - 2p + 1)]/2 )
then x = a(2p-1)
rearranging to make p the subject
p= (x+a)/2a</span>
(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.
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(b) The following functions are positive in quadrant III:
tangent, cotangent
The following functions are negative in quadrant III
cosine, sine, secant, cosecant
A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.
The slope is -2. this is because if you look at the graph it goes down 2 units and it goes right to the once so it’ll be -2/1 which equals -2.
