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3 years ago

5

True or false:

2 answers:

Irina18 [472]3 years ago

8 0

1) True. Make a list of the first few perfect squares (1, 4, 9, 16, 25, 36, 49, 64,81)

2) False. (a + b)(a + b) = a² + 2ab + b²

Blababa [14]3 years ago

3 0

number 1 is true the difference between two consecutive perfect square numbers is always odd

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The University of Central Florida is planning to build a new access ramp for wheelchairs at its education building. The main ent

forsale [732]

Answer:

0.6 feet

Step-by-step explanation:

We solve this question using the Trigonometric function of Tangent.

tan θ = Opposite/Adjacent.

Where:

Opposite = Height /Length of the ramp = ?

Adjacent = Distance from the base of the ramp = 4 feet

θ = 8.1°

Therefore,

tan 8.1° = x/4

Cross Multiply

x = tan 8.1 × 4

x = 0.5692843028 feet

Approximately = 0.6 feet

Length of the ramp = 0.6 feet.

6 0

2 years ago

In the diagram, △ABC∼△DEF. The area of △ABC is 180 square centimeters. Find the area of △DEF.

Alexeev081 [22]

Answer:

20

Step-by-step explanation:

The first step is to find out the similarity reason of the sides of the triangle, so we just divide the big side by the small side (12/4=3).

Now, for the similarity reason of the areas, we just square that number (3^2=9).

Then we just divide the area of the big triangle by 9, which is 180/9= 20.

4 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

3 years ago

Matthew manages the packaging department of his company. Yesterday, the department operated for only 4 hours due to a company pi

11111nata11111 [884]

Answer:

add them all otherwise I don't know im on that question sorry

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

HELP ASAP pleaseeee

AlekseyPX

$\\ \sf\longmapsto 3y=z-2x$

Take z to left

$\\ \sf\longmapsto -2x=3y-z$

Take out -

$\\ \sf\longmapsto 2x=-3y+z$

$\\ \sf\longmapsto 2x=z-3y$

$\\ \sf\longmapsto x=\dfrac{z-3y}{2}$

5 0

3 years ago

Read 2 more answers