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Tems11 [23]
3 years ago
5

An aquarium with a square base has no top. There is a metal frame. Glass costs 10 dollars/m^2 and the frame costs 9 dollars/m. T

he volume is to be 20 m^3. Express the total cost C in terms of the height h in meters. (Hint: work out the cost of the glass and frame separately.)
Mathematics
1 answer:
LenKa [72]3 years ago
3 0
The volume is given as V = Lwh = 20 m^3.
Since the base is a square then V = L²h = 20 m^3
Rearranging, we have L = √(20/h).

For the metal frame:
8√(20/h) + 4h ==> [8 is multiplied to √(20/h) because there are 8 sides measuring L, and 4 is multiplied to h because there are 4 sides that have measure h.]

9[8√(20/h) + 4h]
=72√(20/h) + 36h = Cost of metal frame

For the glass:
Area of one square base = L² = 20/h
Area of one rectangular side = h√(20/h) 
Area of 4 rectangular sides = 4h√(20/h) 

10[20/h + 4h√(20/h)] = Cost of glass

C = 72√(20/h) + 36h + 10[20/h + 4h√(20/h)] 

(take note, we did not multiply 10 to the terms because 10 dollars is per unit area and not unit length.)
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Determine which of the sets of vectors is linearly independent. A: The set where p 1(t) = 1, p 2(t) = t 2, p 3(t) = 1 + 5t B: Th
denis-greek [22]

Answer:

(A)A and C

Step-by-step explanation:

In each case, represent each of the p_i as a column vector where  each row corresponds to the constant term, coefficient of t and t^2 respectively.

A= The set where p_1(t)=1 p_2(t)=t^2 p_3(t)=1+5t

A=\left[\begin{array}{ccc}1&0&1\\0&0&5\\0&1&0\end{array}\right]

|A|=\left|\begin{array}{ccc}1&0&1\\0&0&5\\0&1&0\end{array}\right|=-5

B: The set where p_1(t)=t, p_2(t)=t^2, p_3(t)=2t+5t^2

B=\left[\begin{array}{ccc}0&0&0\\1&0&2\\0&1&5\end{array}\right]\\|B|=0

C: The set where p_1(t)=1, p_2(t)=t^2, p_3(t)=1+5t+t^2

C=\left[\begin{array}{ccc}1&0&1\\0&0&5\\0&1&1\end{array}\right]\\|C|=-5

Since the determinants of A and C are not 0, the set of vectors in A and C are linearly independent.



6 0
3 years ago
Hmm whats the answer?
ehidna [41]
I think x equals 155 I may not be right but that's what I got.
6 0
4 years ago
Read 2 more answers
Does anybody understand writing equations in slope intercept form?
aleksandr82 [10.1K]

slope intercept form

y=mx+b

where m is the slope and b is the y intercept

if we change from point slope form

y-y1 = m(x-x1)

we distribute

y-y1 = mx -x*x1

then add y1 to each side

y = mx -x*x1+y1


remember x and y are variables and should stay in the equation

m,x1,y1 are numbers from the problem

you may have to calculate the slope (m) from the formula

m = (y2-y1)/(x2-x1)  from two points on the line

4 0
3 years ago
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Bob's monthly bill is made up of a $10 fee plus $0.05 per minute. Write an expression that represents Bob's monthly phone if he
Salsk061 [2.6K]

Answer:

10 + 0.05m; m = 350 minutes

Bob's phone bill will be $27.50

Step-by-step explanation:

$10 + ($0.05 x 350 minutes) = $27.50

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B
riadik2000 [5.3K]

Let

S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives

r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}

and subtracting this from S_n gives

(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}

As n → ∞, the exponential term will converge to 0, and the partial sums S_n will converge to

\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}

We must have |r| < 1 since both sums converge, so

\dfrac{15}a = \dfrac1{1-r}

\dfrac{150}{a^2} = \dfrac1{1-r^2}

Solving for r by substitution, we have

\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)

\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}

Recalling the difference of squares identity, we have

\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

It follows that

\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0
2 years ago
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