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3 years ago

Cindy's age, x, is 3 times her age 6 years ago which equation represents the statement x=3(x-6) or x=3-6x or x=3x-6 or x=3x+6

2 answers:

8 0

Let's denote Cindy's age which is unknown with X
We know that Cindy's age is 3 times her age 6 years ago.
6 years ago Cindy's age was the present age (X) - 6 : X-6
The fact that the present age is 3 times her age 6 year ago can be written as:
3*(X-6)
So, finally:
X=3*(X-6)

Lerok [7]3 years ago

5 0

X=3x+6
................

x= 9x

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nlexa [21]

$202 + 156 = 358$
$500 - 358 = 142$
the answer is A. 142

7 0

3 years ago

The length of a rectangle is one more that three times its width. If the perimeter is 42 inches, find the dimensions of the rect

Liono4ka [1.6K]

Let w=width
L= length
3w+1=l
Perimeter -2(l+w)
2(W+3w+1)=42
Divide both sides by 2
W+3w+1=21
Subtract 1
W+3w=20
Add like terms
4w=20
Divide both sides by 4
W=5
The width is 5 inches

3w+1=l
Input the width into the equation
3(5)+1=l
15+1=l
16=l
The length of the rectangle is 16 inches

The rectangle is 5 x 16 inches

6 0

3 years ago

What is the midpoint of EC ?

sergij07 [2.7K]

<u>Given</u>:

Given that the graph OACE.

The coordinates of the vertices OACE are O(0,0), A(2m, 2n), C(2p, 2r) and E(2t, 0)

We need to determine the midpoint of EC.

<u>Midpoint of EC:</u>

The midpoint of EC can be determined using the formula,

$Midpoint=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Substituting the coordinates E(2t,0) and C(2p, 2r), we get;

$Midpoint=(\frac{2t+2p}{2},\frac{0+2r}{2})$

Simplifying, we get;

$Midpoint=(\frac{2(t+p)}{2},\frac{2r}{2})$

Dividing, we get;

$Midpoint=(t+p,r)$

Thus, the midpoint of EC is (t + p, r)

Hence, Option A is the correct answer.

7 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago

What is 8/9 closer to? A. 1/2 B. 1 or C. 0

nikitadnepr [17]

9/9 = 1 and 8/9 is only 1/9 less than 9/9 which is 1.

Answer: B) 1 8/9 is closer to 1

5 0

3 years ago

Read 2 more answers