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3 years ago

If (2 − 3i) + (x + yi) = 6, what is x + yi?

1 answer:

5 0

(2 - 3i) + ( x + yi) = 6
(x + yi) = 6 - (2 - 3i)
Distribute the - to the 2 - 3i and get
x + yi = 6 - 2 + 3i
Accumulate like terms.
x + yi = 4 + 3i. Answer

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Josh has a rewards card for a movie theater. He receives 15 points for becoming a rewards card holder. He earns 3.5 points for e

Savatey [412]

Answer:

C. 55 $\leq$ 3.5x + 15

Step-by-step explanation:

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3 years ago

Carl offered to pay jake $0.25 a square foot to paint his fence. The fence measures 109 feet long and 7 feet wide what is the to

Scorpion4ik [409]

Answer:

Part 1) The total square footage of the fence is $763\ ft^{2}$

Part 2) $\$190.75$

Step-by-step explanation:

Part 1) what is the total square footage of the fence ?

we know that

The area of the fence is equal to

$A=LW$

substitute the values

$A=(109)(7)=763\ ft^{2}$

Part 2) How much will Jake get paid

Using proportion

$\frac{0.25}{1}\frac{\$}{ft^{2}}=\frac{x}{763}\frac{\$}{ft^{2}}\\ \\x=763*0.25\\ \\x=\$190.75$

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3 years ago

Find the linear function with the following properties. <br><br> f(0)=−9 Slope of f=−8

Ratling [72]

Answer:

f(x) = -8x - 9

Step-by-step explanation:

<h3>Given </h3>

<u>Function f, with </u>

f(0) = -9
Slope of f = -8

Linear function

<h3>Solution</h3>

<u>Slope-intercept formula</u>

f(x) = mx + b
f(0) = -9 means y-intercept b = -9, and the slope is m = - 8

<u>So the function is</u>

f(x) = -8x - 9

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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.