Let A = 0, 1 , B = 2, 3, 4 and C = 3, 5 verity that A x (BUC) = (AXB) U (AXC)
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Answer:

<CAB =30°

Step-by-step explanation:

<DAB=80°

<DAC=50°

<DAB=<DAC+<CAB

<CAB=<DAB-<DAC

=80°-50°

=30°

Assuming 120 is a base 10 numeral, then

120 = 100 + 20 = 4×5² + 4×5¹

so you have

120₁₀ = 440₅

Answer: $4.50

Step-by-step explanation:

48-21=27

27 divided by 6= 4.50

Therefore each case costs 4.50

Answer:

$4.50

Step-by-step explanation:

48 - 21

27

divide by 6

4.50

Answer:

D. 10

Step-by-step explanation:

[tex]{ \sf{(12x + 12) \degree = 79\degree + (5x + 3)\degree}} \\ { \sf{ \{outer \: angle \: is \: equal \: to \: sum \: of \: inner \: angles \}}} \\ { \sf{12x - 5x = 79 + 3 - 12}} \\ { \sf{7x = 70}} \\ { \sf{x = 10}}[/tex]

Answer:

position of objects in space

for example: up, down, in front, behind, between, left, right.

motion of objects in space

for example: Stars, planets, moons.

Answer:

17) 19

19) 8

Step-by-step explanation:

17)

(6 +6² -4)÷2

(6 +36 -4)÷2

(38)÷2

19

19)

3³ ÷3 - 1

27 ÷3 -1

9 -1

8

Answer:

null set

Step-by-step explanation:

i am not sure but i hope it will help you

Answer:

singleton set is correct

Hi ;-)

[tex]y-4=3\\\\y=3+4\\\\\boxed{y=7}[/tex]

Answer:  7  

.....7-4=3

Answer:

[tex]{ \underline{ \sf{6y = x + 1 }}}[/tex]

Step-by-step explanation:

[tex]y = mx + c[/tex]

m is slope

c is y intercept.

at x-intercept, y = 0

[tex]0 =( \frac{1}{6} \times - 1 )+ c \\ \\ c = \frac{1}{6} [/tex]

A line is a one-dimensional shape that is straight, has no thickness, and extends in both directions indefinitely. The equation of the given line is 6y=x+1.

A line is a one-dimensional shape that is straight, has no thickness, and extends in both directions indefinitely. The equation of a line is given by,

y =mx + c

where,

x is the coordinate of the x-axis,

y is the coordinate of the y-axis,

m is the slope of the line, and

c is the y-intercept.

Given that the slope of the equation of the line is 1/6, while the x-intercept is at -1. Therefore, if we substitute the coordinate of the intercept and the slope in the equation in the general equation of the line. We will see,

y = mx + c

0 = (1/6)(-1) + c

0 = -1/6 + c

c = 1/6

Now, the equation of the line can be written as,

y = (1/6)x + (1/6)

y = (x/6) + (1/6)

y = (x+1)/6

6y = x + 1

Hence, the equation of the given line is 6y=x+1.

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Answer:

If the exponent is 0 the answer is 224/y^2

Step-by-step explanation:

Answer:

1/23

Step-by-step explanation:

5/23 - 4/23 would be 1/23. When the denominators of the fractions are the same, you can just subtract the numerators.

Answer:

Step-by-step explanation:

[tex] \frac{3}{5} xy \: \times ( - 25 {x}^{2} y)[/tex]

[tex] = \frac{3}{5} \times ( - 25) \times x \times {x}^{2} \times y \times y[/tex]

[tex] = - 15 \times {x}^{3} \times {y}^{2} [/tex]

[tex] = - 15 {x}^{3} {y}^{2} (ans)[/tex]

Hi ;-)

[tex]s=zh-2zt^3\\\\z(h-2t^3)=s \ \ /:(h-2t^3)\\\\z=\dfrac{s}{h-2t^3}[/tex]

Answer:

z = [tex]\frac{s}{h-2t^3}[/tex]

Step-by-step explanation:

Given

s = zh - 2zt³ ← factor out z from each term

s = z(h - 2t³ ) ← divide both sides by h - 2t³

[tex]\frac{s}{h-2t^3}[/tex] = z

Answer:

Hope it helped

Step-by-step explanation:

The restricted domain that makes the given function one to one is (-∞,0] or [0.∞) and the inverse function  would be f⁻¹ (x) = [tex]\sqrt{x-2}[/tex] if we restrict the domain of original function to [0.∞) and the inverse function would be  f⁻¹ (x) = -[tex]\sqrt{x-1}[/tex] if we restrict the domain to (-∞,0] .

An inverse function or an anti function is defined as a function, which can reverse into another function.

Now the given quadtratic function is,

f(x) = x² + 2

Now vertex of the given quadratic function is (0,2)

Thus the range of the function is [2,∞)

If we restrict the domain of this function to either (-∞,0} or [0.∞), it will become one to one function.

Now the inverse function is given as,

y = x² + 2

Upon interchanging x and y, we get:

x = y² + 2

⇒ y² = x - 2

or, y = ±[tex]\sqrt{x-2}[/tex]

Hence, the inverse function would be f⁻¹ (x) = [tex]\sqrt{x-2}[/tex] if we restrict the domain of original function to [0.∞) and the inverse function would be  f⁻¹ (x) = -[tex]\sqrt{x-1}[/tex] if we restrict the domain to (-∞,0] .

Thus, the restricted domain that makes the given function one to one is (-∞,0] or [0.∞) and the inverse function  would be f⁻¹ (x) = [tex]\sqrt{x-2}[/tex] if we restrict the domain of original function to [0.∞) and the inverse function would be  f⁻¹ (x) = -[tex]\sqrt{x-1}[/tex] if we restrict the domain to (-∞,0] .

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Answer:

600

Step-by-step explanation:

since 1m=1000mm

Answer:

40

Step-by-step explanation:

137+143=280

280/7= 40

Answer:

Division/dividing/divide

Step-by-step explanation:

If Justin has 280 baseball cards, and if each page will only hold 7 cards per page, then by dividing 280 by 7 (280÷7) would equal to 40 pages.

Answer:

Step-by-step explanation:

In x², here the 2 is exponent while x is base.

Answer:

an exponent

Step-by-step explanation:

hope it helps:)

Answer:

Step-by-step explanation:

[tex]U=\{(x,y)\ \zeta\ y=|x+1| \ and\ x \in \{-2,-1,0,1,2\}\ \}\\\\\begin{array}{|c|c|}x&y\\---&---\\-2&2\\-1&0\\0&2\\1 & 4\\2&6\\---&---\\\end{array}\\\\U=\{(-2,2),(-1,0),(0,2),(1,4),(2,6)\}\\[/tex]

Answer:

2 / (-2,2)

Step-by-step explanation:

Answer:

x = 17°

m∠MKL = 116°

m∠JKL = 159°

Step-by-step explanation:

JKM = 43

MKL = 8x - 20

JKL = 10x - 11

Since JKL is the whole angle, JKM + MKL = JKL

43 + 8x - 20 = 10x - 11

Add 20 to both sides:

43 + 8x = 10x + 9

Subtract 8x from both sides:

43 = 2x + 9

Subtract 9 from both sides:

34 = 2x

x = 17°

MKL = 8(17) - 20

MKL = 116°

JKL = 10(17) - 11

JKL = 159°

And to check 159 - 116 = 43

If my answer is incorrect, pls correct me!

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-Chetan K

Answer:

[tex]\frac{(a+b)}{a^{2} -b^{2} }[/tex] + [tex]\frac{(a-b)}{a^{2} -b^{2} }[/tex] = [tex]\frac{2a}{a^{2} -b^{2} }[/tex]

Step-by-step explanation:

Step-by-step explanation:

[tex] \frac{ \sqrt{a + b} }{ \sqrt{a - b} } + \frac{ \sqrt{a - b} }{ \sqrt{a + b} } \\ squaring \: on \: both \: sides \\ \frac{a + b}{a - b} + \frac{a - b}{a + b} \\ \frac{(a + b) \times (a + b) + (a - b)(a - b)}{(a - b)(a + b)} \\ \frac{ {(a + b)}^{2} + {(a - b)}^{2} }{ {a}^{2} - {b}^{2} } \\ remaining \: in \: attachment[/tex]

[tex]thank \: you[/tex]

Answer:

8

Step-by-step explanation:

The function u given me is

[tex] \tan( \frac{\pi}{4} (x - \frac{\pi}{3} ) )[/tex]

The period of a function is represented by

[tex] \frac{2\pi}{ |b| } [/tex]

where b is the coefficient of the x variable.

The coefficient is pi/4 so

[tex] \frac{2\pi}{ \frac{\pi}{4} } = 8[/tex]

8 is the answer.

Answer: 4 units

Step-by-step explanation:

Answer:

[tex] \sqrt{2.2.3.3.5} [/tex]

[tex]\sqrt{2.2.3.3.5}[/tex] is equivalent to [tex]\sqrt{180}[/tex]. Correct option is 1. First, we look for perfect square factors within 180. We can factor 180 as follows: [tex]\(180 = 2^2 \times 3^2 \times 5\)[/tex]

Next, we can group the perfect square factors together:

[tex]\(180 = (2 \times 3)^2 \times 5\)[/tex]

Since we are taking the square root of 180, we can rewrite it as the square root of its perfect square factors multiplied by the remaining non-perfect square factor:[tex]\(\sqrt{180} = \sqrt{(2 \times 3)^2 \times 5}\)[/tex]

The square root of a perfect square is just the value of the number itself:

[tex]\(\sqrt{180} = 2 \times 3 \times \sqrt{5}\)[/tex]

Finally, we can simplify the expression by multiplying the numbers outside the square root [tex]\(\sqrt{180} = 6\sqrt{5}\)[/tex]

So, the expression [tex]\(6\sqrt{5}\)[/tex] is equivalent to [tex]\(\sqrt{180}\)[/tex], and it represents the positive square root of 180 with any perfect square factors taken outside the square root sign.

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Area=50m^2

Let side be a

[tex]\\ \rm\Rrightarrow Area=a^2[/tex]

[tex]\\ \rm\Rrightarrow a^2=50[/tex]

[tex]\\ \rm\Rrightarrow a=\sqrt{50}[/tex]

[tex]\\ \rm\Rrightarrow a=7.1m[/tex]

Answer:

[tex]\displaystyle 5\sqrt{2}\:m.[/tex]

Step-by-step explanation:

Sinse squares are equal in length all around, this formula applies:

[tex]\displaystyle s^2 = A → \sqrt{50} = \sqrt{s^2} \\ \\ 5\sqrt{2} = s[/tex]

Therefore, the length of the playground is [tex]\displaystyle 5\sqrt{2}[/tex] metres.

* Just in case you could not figure out how to find the square root of fifty, here is how using perfect squares:

[tex]\displaystyle \sqrt{50} → \sqrt{2 \times 25} \\ \\ 5\sqrt{2}[/tex]

Sinse we are dealing with measurements, we only want the NON-NEGATIVE root.

I am joyous to assist you at any time.

Answer:

I think its c I hope I'm right

The orthogonal decomposition of vector [tex]\vec b = <9, 0, 0>[/tex] with respect to vector [tex]\vec a = <4, -5, 0>[/tex]  is

[tex]b'=<\frac{16}{5},\frac{8}{5},0>[/tex]

From the Question we are told that

Vector [tex]\vec b = <9, 0, 0>[/tex]

Vector [tex]\vec a = <4, -5, 0>[/tex]

Generally in the orthogonal decomposition of b to a we have

[tex]\vec b=\vec b"+\vec b'[/tex]

Where

[tex]\vec b"=(\frac{\vec b*\vec a}{\vec *\vec a})*\vec a[/tex]

[tex]\vec b"=(\frac{ <9, 0, 0>*<4, -5, 0>}{<4, -5, 0>*<4, -5, 0>})*<4, -5, 0>[/tex]

[tex]\vec b"=<\frac{4}{5},\frac{-8}{5},0>[/tex]

Therefore

[tex]b'= \vec b- \vec b"\\\\b'=<4,0,0>-<\frac{4}{5},\frac{-8}{5},0>[/tex]

[tex]b'=<\frac{16}{5},\frac{8}{5},0>[/tex]

in Conclusion

The orthogonal decomposition of vector [tex]\vec b = <9, 0, 0>[/tex] with respect to vector [tex]\vec a = <4, -5, 0>[/tex]  is

[tex]b'=<\frac{16}{5},\frac{8}{5},0>[/tex]

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Answer:

measure of both angle is 66

Step-by-step explanation:

vertical angles are equal so,

6x+15=4x+32

6x-4x=32-15

2x= 17

X=8.5

now,

4X+32

= 4×8.5 +32

= 66

Answer:

40+x+5 = 3x +15

2x= 30

x=15

Their present ages are 15 and 55

Answer:

-14°F

Step-by-step explanation:

Hey there!