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Keith_Richards [23]

3 years ago

8

Value of the digit 3 in 20,320

2 answers:

Tema [17]3 years ago

8 0

I believe that the answer is 300

vazorg [7]3 years ago

7 0

The answer is hundred

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Use the divergence theorem to find the outward flux of the vector field F(x,y,z)=2x2i+5y2j+3z2k across the boundary of the recta

aksik [14]

Answer:

The answer is "120".

Step-by-step explanation:

Given values:

$F(x,y,z)=2x^2i+5y^2j+3z^2k \\$

differentiate the above value:

$div F =2 \frac{x^2i}{\partial x}+5 \frac{y^2j}{\partial y}+3 \frac{z^2k }{\partial z} \\$

$= 4x+10y+6z$

$\ flu \ of \ x = \int \int div F dx$

$= \int\limits^1_0 \int\limits^3_0 \int\limits^1_0 {(4x+10y+6z)} \, dx \, dy \, dz \\\\ = \int\limits^1_0 \int\limits^3_0 {(4xz+10yz+6z^2)}^{1}_{0} \, dx \, dy \\\\ = \int\limits^1_0 \int\limits^3_0 {(4x+10y+6)} \, dx \, dy \\\\ = \int\limits^1_0 {(4xy+10y^2+6y)}^3_{0} \, dx \\\\ = \int\limits^1_0 {(12x+90+18)}\, dx \\\\= {(12x^2+90x+18x)}^{1}_{0} \\\\= {(12+90+18)} \\\\=30+90\\\\= 120$

6 0

3 years ago

Helppppp pleSeeeeeeeee

Zepler [3.9K]

Answer: C

Step-by-step explanation: Hope this help :D

3 0

3 years ago

8 1/2 + 7 2/3 + 4 1/2

SCORPION-xisa [38]

Answer:

62/3 is the answer

hope this helps! :)

Step-by-step explanation:

5 0

3 years ago

Order from least to greatest 3.2 ,3.14, 19/7, sqrt (14)

Oksanka [162]

Answer: 19/7 ,3.14, 3.2

Step-by-step explanation: Hope this helps Have a brilliant day of learning!

7 0

3 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

<em>C.</em> $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em> $(5-\frac{1}{2})^6$ <em />

3 0

3 years ago