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3 years ago

The domain of the function f(x) = √2-x is the set of numbers x satisfying:

Mathematics

1 answer:

jekas [21]3 years ago

7 0

Try this solution:

D(f): √(2-x)≥0; ⇔ 2-x≥0; ⇔ x≤2 or x∈(-∞; 2].

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Nina has a job repairing sailboats. To repair a sailboat, she charges a flat fee of $100 plus $40 per hour. The total in dollars

vekshin1

Answer:

y=100+40x

Step-by-step explanation:

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4 years ago

Sin 30Degrees = square of 3/2 and cos 30degrees =1/2 true or false

zhenek [66]

Answer:

False

Step-by-step explanation:

Trying it in a calculator says otherwise meaning it's false

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3 years ago

The integers that are less than 2 but greater than -4

My name is Ann [436]

The integers or -3, -2, -1,0, and 1

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3 years ago

The diagram shows a parallelogram.

emmasim [6.3K]

Answer:

27.44 square cm

Step-by-step explanation:

If the length of parallelogram is a and b and angle between side a and b is $\alpha$ .

Then area of parallelogram = $a*b*sin(\alpha ) = ab sin(\alpha )$

Given side length 4 cm and 7 cm

angle between them = 100°

value of sin(100°) = 0.98

Thus, area of given parallelogram = 4*7*sin(100°) = 28*0.98 = 27.44

Thus, area of given parallelogram is 27.44 square cm.

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3 years ago

Read 2 more answers

You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you want to place it whe

strojnjashka [21]

Answer:

Step-by-step explanation:

$\text{The equation of the surface of the sphere whose radius is 6 can be represented as:}$

$x^2+y^2+z^2 = 36}$

$So , g(x,y,z) = x^2 + y^2 + z^2 - 36$

$\text{To minimize the function :} \\ \\ M(x,y,z0 = 6x - y^2 +xz + 60$

$\text{by applying lagrange multipliers }$

$\bigtriangledown M (x,y,z) = \lambda \bigtriangledown g(x,y,z)$

$\langle 6+zz,-2y,x \rangle = \lambda \langle 2x,2y,2z \rangle$

$6+z = 2\lambda x$

$-2y = 2 \lambda y$

$\text{from the second equation ;} (\lambda +1)y = 0, \text{so , it is either y =0 or }\lambda = -1$

$Suppose \ \lambda = -1 ; \text{other equation becomes x = -2z and 6+z = -2x}$

$\text{such that; 6+z =4z or z = 2}$

$\text{it implies that: x= -4 and }y = \pm \sqrt{36 - 4-16} = \pm 4$

$\text{Here; there exist two possible points}$

$\text{(-4,-4,2) and (-4,4,2)}$

$\text{In the scenario here y=0,} \lambda \text{ is unknown, then we remove it}$

$x = 2 \lambda z \\ \\ 6+z = 2 \lambda x \\ \\ 6+z = 4 \lambda ^2z \\ \\ 6 = (4 \lambda ^2 -1) z ---(1)$

$z = \dfrac{6}{4 \lambda ^2 -1}$

$x = \dfrac{12 \lambda }{4 \lambda ^2 -1 }$

$\text{recall that; it i possible to divide }4 \lambda ^2 -1 \text{since (1) shows that it cannot be equal to zero.} \\ \\ \text{hence, puttinf this into constraint, whereby y =0, Then:}$

$144 \lambda ^2 + 36 = 36(4\lambda ^2 -1) \\ \\ 4\lambda ^2 ( 4\lambda ^2-3) =0 \\ \\ \lambda = 0 \ or \ \lambda = \pm \dfrac{\sqrt{3}}{2}$

$\text{Thus; the points are:} \\ \\ (0,0,6) \ and \ (\pm 3\sqrt{3},0,3})$

$Now;$

$M(-4,-42) = 12 \\ \\ M(-4,4,2) = 28 \\ \\ M(0,0,6) = 60 \\ \\ M(3\sqrt{3},0,3) = 27\sqrt{3} + 60 \\ \\ M(-3\sqrt{3},0,3) = -27\sqrt{3} +60$

$\text{Hence, the minimum is 12 which occurs at (-4,-4,2)}$

4 0

3 years ago