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lukranit [14]
3 years ago
8

How to put 40,023,032 in expanded form

Mathematics
1 answer:
Valentin [98]3 years ago
5 0
40,023,032 = 40,000,000 + 20,000 + 3,000 + 30 + 2
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4 = 3а - 14<br> How do I solve this problem
Simora [160]

Answer:

X = 6

Step-by-step explanation:

Add 14 to both sides: 18 = 3a

Divide both sides by 3: a = 6

5 0
3 years ago
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Write this statement in your own words: ∃x ∈ ℕ, y ∈ ℤ|x² = y² Rewrite this using the appropriate mathematical notation: Even num
Serga [27]

Answer:

See below.

Step-by-step explanation:

1)

So we have:

\exists x\in\mathbb{N},y\in\mathbb{Z}|x^2=y^2

This can be interpreted as:

"There exists a natural number <em>x</em> and an integer <em>y</em> such that x² is equal to y²."

2)

So we want even numbers are in the set of integers.

\{2n:n\in\mathbb{Z}\}\in\mathbb{Z}

This is interpreted as:

"The set of even numbers (2n such that n is an integer) is in the set of integers"

6 0
3 years ago
In the paper airplane shown , ABCD=EFGH, m
Nookie1986 [14]

Answer:

i would think it would be the same as angle bcd so 90

Step-by-step explanation:

4 0
3 years ago
For each of 2 ≤ n ≤ 5, compute c(n, 0) + c(n, 2) + ⋅⋅⋅ + c(n, 2k), where 2k is largest even integer ≤ n, and compute c(n, 1) + c
Ksivusya [100]
Denote C(n,m)=\dbinom nm. Then

n=2\implies\displaystyle\binom20+\binom22=1+1=2
n=3\implies\displaystyle\binom30+\binom32=1+3=4
n=4\implies\displaystyle\binom40+\binom42+\binom44=1+6+1=8
n=5\implies\displaystyle\binom50+\binom52+\binom54=1+10+5=16

In general, it would appear that

\displaystyle\sum_{k=0}^{2\lfloor\frac n2\rfloor}\binom n{2k}=2^{n-1}

On the other hand,

n=2\implies\displaystyle\binom21=2
n=3\implies\displaystyle\binom31+\binom33=3+1=4
n=4\implies\displaystyle\binom41+\binom43=4+4=8
n=5\implies\displaystyle\binom51+\binom53+\binom55=5+10+1=16

so that in general, we also get

\displaystyle\sum_{k=0}^{2\lfloor\frac{n-1}2\rfloor+1}\binom n{2k+1}=2^{n-1}
7 0
3 years ago
Derive <br><br>Sinxy = x² + 2xy​
bogdanovich [222]

Answer:

    \frac{dy}{dx} = \frac{2x +2 y- ycosxy}{x(cosxy-2)}

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given

        sin xy = x² + 2xy ...(i)

Apply

       \frac{d}{dx} UV = U^{l} V + U V^{l}

     \frac{d}{dx} x^{n} = nx^{n-1}

Differentiating equation (i) with respective to 'x', we get

      cosxy \frac{d}{dx} (xy) = 2x + 2( x\frac{dy}{dx} + y(1))

      cosxy  (x\frac{dy}{dx} +y(1)) = 2x + 2( x\frac{dy}{dx} + y(1))

     cosxy  (x\frac{dy}{dx}) + ycosxy = 2x + 2( x\frac{dy}{dx}) +2 y)

   cosxy  (x\frac{dy}{dx}) - 2( x\frac{dy}{dx}) = 2x +2 y- ycosxy

   (cosxy-2)  (x\frac{dy}{dx})  = 2x +2 y- ycosxy

       \frac{dy}{dx} = \frac{2x +2 y- ycosxy}{x(cosxy-2)}

8 0
3 years ago
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