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Digiron [165]
3 years ago
8

The length of a rectangle is a two-digit number with identical digits (aa). the width is 1/10 of the length. the perimeter is 2

times the area of the rectangle. find the length and the width.
Mathematics
1 answer:
Pavlova-9 [17]3 years ago
5 0
L = 10a + a
L = 11 a
W = 1/10 (11a)
W = 1.1 a

Perimeter = 2(L + W)
Perimeter = 2*(11a + 1.1a)
Perimeter = 22a + 2.2 a
Perimeter = 24.2 a

Area = L * W
Area = 11a * 1.1a
Area = 12.1 a^2
 
The perimeter is 2*area
P = 2*A
P = 2* 12.1 a^2
P = 24.2 a^2

Now put in the perimeter on the left side of the above equation.
24.2a = 24.2 a^2 Divide both sides by 24.2
1 a = a^2 Divide by a
1 = a

The length = 11*a = 11*1 = 11
The Width =  1.1 a = 1.1 * 1 = 1.1

Answers
L = 11 <<<<<<< answer
W = 1.1 <<<<<< answer
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Eight more than x is at least fourteen. what is the solution set for x?
Otrada [13]
X + 8 >= 14
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3 years ago
How do you factor 35x^2+26-16? Show work please.
Sedbober [7]

well, first off, we do a quick prime factoring of the leading term's coefficient, 35, and the constant, 16.

35 = 5*7

16 = 2*2*2*2

and the rest is just a mix and match.

what combination from the factors of 35 and the factors of 16, can provide a product whose sum is the middle term 26?  well, without further adieu.


we can try 5*2*2  + 7*2*2, and that's 20 and 28, ok, 28 - 20 is 8, so that won't work.

another can be 5*2*7  and 2*2, that's 70 and 4, 70 - 4 is 56, that won't work either.

5*2*2*2 and 7*2, that's 40 and 14, hmmm 40 - 14 = 26!!! holly guacamole!

so then,


\bf 35x^2+26x-16\implies (5x-\stackrel{\stackrel{2}{\downarrow }}{2})(7x+\stackrel{\stackrel{2\cdot 2\cdot 2}{\downarrow }}{8})

3 0
4 years ago
This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex
pashok25 [27]

Answer with Step-by-step explanation:

We are given that

f(x,y,z)=10x+10y+3z

g(x,y,z)=5x^2+5y^2+3z^2=43

We have to find the extreme values of the function.

f_x(x,y,z)=10,f_y(x,y,z)=10,f_z(x,y,z)=3

g_x(x,y,z)=10x,g_y(x,y,z)=10y,g_z(x,y,z)=6z

\Delta f(x,y,z)=\lambda \Delta g(x,y,z)

Therefore,

10=\lambda 10x\implies x=\frac{1}{\lambda}

10=10y\lambda\implies y=\frac{1}{\lambda}

3=6z\lambda

z=\frac{1}{2\lambda}

Substitute the values in g(x)

Then, we get

\frac{5}{\lambda^2}+\frac{5}{\lambda^2}+\frac{3}{4\lambda^2}=43

\frac{20+20+3}{4\lambda^2}=43

\frac{43}{4\lambda^2}=43

\lambda^2=\frac{43}{4\times 43}

\lambda=\pm\frac{1}{2}

When \lambda=\frac{1}{2} then,

x=2,y=2,z=1

When \lambda=-\frac{1}{2}

Then, x=-2,y=-2,z=-1

f(2,2,1)=10+10+3=23

f(-2,-2,-1)=-10-10-3=-23

Hence, maximum value of f(x,y,z) is 23 at (2,2,1) and minimum value of f(x,y,z) is -23 at (-2,-2 ,-1).

8 0
4 years ago
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