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cupoosta [38]
1 year ago
9

Graph the six terms of a finite sequence where a1 = 5 and r = 1.25

Mathematics
1 answer:
Karo-lina-s [1.5K]1 year ago
5 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a1 = 5 
<span>a2 = 5*r = 5*(5/4) = 25/4 </span>
<span>a3 = 5*r² = 5*(5/4)² = 125/16 </span>
<span>a4 = 5*r³ = 5*(5/4)³ = 625/256 </span>
<span>a5 = 5*r⁴ = 5*(5/4)⁴ = 3125/1024</span>

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Expand (2x-3y)^4 using Pascal's Triangle. Show work
deff fn [24]
(2x+3y)⁴
1) let 2x = a   and 3y = b

(a+b)⁴ = a⁴ + a³b + a²b² + ab³ + b⁴
Now let's find the coefficient of each factor using Pascal Triangle
     
                     0     |               1
                     1     |            1    1
                     2     |          1   2   1
                     3     |         1  3   3   1
                     4     |       1  4   6    4  1

0,1,2,3,4,.. represent the exponents of binomials 
Since our binomial has a 4th exponents, the coefficients are respectively:

(1)a⁴ + (4)a³b + (6)a²b² + (4)ab³ + (1)b⁴
Now replace a and b by their real values in (1):

2⁴x⁴ +(4)8x³(3y) + (6)(2²x²)(3²y²) + (4)(2x)(3³y³) + (1)(3⁴)(y⁴)

16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴
3 0
2 years ago
A number k is less than 3 units from 10. <br> What's the absolute value inequality???
Brrunno [24]

It would be: k+3 = 10

subtracting 3 from both sides,

k+3-3 = 10-3

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6 0
2 years ago
Find the point (x,y) of x2+14xy+49y2=100 that is closest to the origin and lies in the first quadrant.
Gala2k [10]

Notice that

x^2+14xy+49y^2=(x+7y)^2

so the constraint is a set of two lines,

(x+7y)^2=100\implies\begin{cases}x+7y=10\\x+7y=10\end{cases}

and only the first line passes through the first quadrant.

The distance between any point (x,y) in the plane is \sqrt{x^2+y^2}, but we know that \sqrt{f(x,y)} and f(x,y) share the same critical points, so we need only worry about minimizing x^2+y^2. The Lagrangian for this problem is then

L(x,y,\lambda)=x^2+y^2+\lambda(x+7y-10)

with partial derivatives (set equal to 0)

L_x=2x+\lambda=0

L_y=2y+7\lambda=0

L_\lambda=x+7y-10=0

We have

L_y-7L_x=2y-14x=0\implies y=7x

which tells us that

x+7y-10=0\iff x+49x=10\implies x=\dfrac15\implies y=\dfrac75

so that \left(\dfrac15,\dfrac75\right) is a critical point. The Hessian for the target function x^2+y^2 is

H(x,y)=\begin{bmatrix}2&0\\0&2\end{bmatrix}

which is positive definite for all x,y, so the critical point is the site of a minimum. The minimum distance itself (which we don't seem to care about for this problem, but we might as well state it) is \sqrt{\left(\dfrac15\right)^2+\left(\dfrac75\right)^2}=2.

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2 years ago
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The answer is A I think
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Furkat [3]

Answer:

The first step is to multiply by a power of 10, so the divisor is a whole number.

Step-by-step explanation:

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2 years ago
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