Ask question

Ask question

All categories

sleet_krkn [62]

2 years ago

11

Plzz help i been asking this question for way too long

1 answer:

kakasveta [241]2 years ago

7 0

Step-by-step explanation:

we know that that the stickers are 1 inch long and 1 inch wide.

We also know that 1x1=1

There are also 3 stickers.

1x3=3

3 is your area.

You might be interested in

Let A = {10, 20, 30, 40, 50, 60} and B = {10, 40, 60}. What is A ∩ B?

erastovalidia [21]

This means common members {10,40,60}

5 0

2 years ago

Read 2 more answers

Solve this inequality: -3 - 6(4x + 6) > -111

Semenov [28]

Step-by-step explanation:

-3-24x-36>-111

-39-24x>-111

-24x>-111+39

-24x>-72 /x(-1)

24x<72

x<2.66

5 0

3 years ago

Hal is asked to write an exponential function to represent the value of a $10,000 investment decreasing at 2% annually. What

jek_recluse [69]

Answer:

0.98

Step-by-step explanation:

The investment starts at $10,000.

$10,000 is 100% of the investment.

In the first year, it loses 2% of its value. That means that after 1 year, the value of 100% of $10,000 is decreased by 2% of $10,000.

100% - 2% = 98%

After the first year, the investment is worth 98% of the original value.

98% as a decimal is 0.98

Answer: 0.98

4 0

3 years ago

Read 2 more answers

Maria does weightlifting for 30 seconds and then aerobics for 60 seconds. She repeats this 10 times. If she does not take any br

VikaD [51]

Answer:

15 minutes

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

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