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2 years ago

8

Forty is no greater than the difference of a number and 2

1 answer:

nikklg [1K]2 years ago

8 0

40≤n-2 is the formula

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(07.03 LC) The steps below show the incomplete solution to find the value of q for the equation 4q − 2q − 4 = −1 + 21: Step 1: 4

Elza [17]

Answer:

3: 2q − 4 = 20

Step-by-step explanation:

Solve for q:

2 q - 4 = 20

Add 4 to both sides:

2 q + (4 - 4) = 4 + 20

4 - 4 = 0:

2 q = 20 + 4

20 + 4 = 24:

2 q = 24

Divide both sides of 2 q = 24 by 2:

(2 q)/2 = 24/2

2/2 = 1:

q = 24/2

The gcd of 24 and 2 is 2, so 24/2 = (2×12)/(2×1) = 2/2×12 = 12:

Answer: q = 12

8 0

2 years ago

Read 2 more answers

Which of the following shows the correct steps to find the value of 27 to the power of 1 over 3?

kramer

Anything to the power of one is it self so
27^1/3
27/3
9

5 0

2 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

2 years ago

While preparing his budget sheet richard Smith noticed that his annual expenses were 18800 while his monthly living expenses wer

PIT_PIT [208]

Answer:

Monthly Living Expense = 2375

Monthly Fixed Expense = 1975

Step-by-step explanation:

Let monthly living expense be "x"

and monthly fixed expense be "y"

<u>"monthly living expenses were 400 more than his monthly fixed expenses":</u>

x = 400 + y

<u>"Richard total monthly expenses were 4350":</u>

x + y = 4350

Putting Equation 1 in Equation 2:

(400 + y) + y = 4350

400 + 2y = 4350

2y = 3950

y = 1975

Now, x = 400 + y = 400 + 1975 = 2375

8 0

3 years ago

Which symbol should go in the blank<br> 8x 5/4 __8

mestny [16]

Answer:

it should be x

Step-by-step explanation:

hope this helps

3 0

2 years ago