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2 years ago

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Simplify: –3(y 2)2 – 5 6y

1 answer:

aleksandr82 [10.1K]2 years ago

8 0

The answer is -3y^4+6y-5

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Verda used a sensor to measure the speed of a moving car at different times. At

Kaylis [27]

Answer:

the answer is c because if you divided 4848484 by 37 =234

Step-by-step explanation:

393-=3\3

3 0

2 years ago

Write an equation in point-slope form and in slope-intercept form.

Sever21 [200]

The point-slope form:

$y-y_1=m(x-x_1)$

The slope-intercept form:

$y=mx+b$

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (1, -5) and (-3, -7). Substitute:

$m=\dfrac{7-(-5)}{-3-1}=\dfrac{12}{-4}=-3\\\\y-(-7)=-3(x-(-3))\\\\\boxed{y+7=-3(x+3)}$

Rewrite to the slope-intercept form:

$y+7=-3(x+3)$ <em>use distributive property</em>

$y+7=-3x-9$ <em>subtract 7 from both sides</em>

$\boxed{y=-3x-16}$

3 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

Find the 76th term of the arithmetic sequence 17 19 21

OLga [1]

Answer:

167

Step-by-step explanation:

4 0

2 years ago

What is the perimeter and area for the glowing figure?

nadya68 [22]

Answer:

Area=15

Perimeter=16

Step-by-step explanation:

First, section the shape off by easier work with shapes.

The two end triangles can then form one large rectangle that is 5in. x 3in.

Next just multiply the length times the width for the area which is 15 in., and add all of the sides up, which will give you 16 in.

8 0

1 year ago