Answer:
0.25
Step-by-step explanation:
Average means 
According to the formula, each student in Mr. Goldwasser's class read

books on average, and each student in Mr. Welch's class read

books on average.
Therefore, the answer will be

Answer: x = 9.6
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Explanation:
We have two smaller right triangles that are glued together so to speak.
The base of the smaller triangle on the left is 5 while the height is h.
Let's use the tangent rule to find the value of h
tan(angle) = opposite/adjacent
tan(55) = h/5
5*tan(55) = h
h = 5*tan(55)
h = 7.14074003371058
Make sure your calculator is in degree mode. That value of h above is approximate.
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Now focus on the smaller triangle on the right
It has the same height value h. This side is the adjacent side while x is the hypotenuse.
We'll use the cosine ratio
cos(angle) = adjacent/hypotenuse
cos(42) = h/x
cos(42) = 7.14074003371058/x
x*cos(42) = 7.14074003371058
x = 7.14074003371058/cos(42)
x = 9.6088135029715
x = 9.6
Answer:
Sorry to not actually answer the question, you can just ignore me but no one can answer since you didnt give the full problem, which would be the measurments of the project etc.
Answer: $76,244.51
Step-by-step explanation:
You need to use the compound interest formula here.
First of all however, you need to convert the terms to monthly figures because the interest is compounded monthly.
4% in months = 4 / 12 = 4/12%
6 years = 6 * 12 = 72 months
Now use the compound interest formula:
= Amount * (1 + rate) ^ number of years
= 60,000 * ( 1 + 4/12%) ⁷²
= $76,244.51
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)
We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>
The vector product pq x pr gives a vector perpendicular to both pq and pr. This vector is the normal vector of a plane passing through all three points
pq x pr
=
i j k
-4 -2 -4
-3 5 1
=<-2+20,12+4,-20-6>
=<18,16,-26>
Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>
The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).
The equation of the required plane is therefore
Π : 9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π : 9x+8y-13z=24
Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.