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2 years ago

Find domain of this function. What you see are two rays on the graph.:

Mathematics

1 answer:

taurus [48]2 years ago

8 0

The correct answer is C
Hope this helped :)

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Simplify the following expression. log(x2) - log(x)

nadya68 [22]

Bc. log a - log b = log (a/b) than use this formula will get

log x^2 -log x = log (x^2 /x) = log x so choice D. is right sure

hope helped

6 0

3 years ago

A=lw+wh+lh, solve for w

KIM [24]

Subtract what is to the right of the equal sign form both side of the equation but any way the answer is 0

6 0

2 years ago

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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

2 years ago

26. 50 students sit in a class room. Each student

Gnom [1K]

<h2><em>each student requires 9m² of floor </em></h2><h2><em>and given the no. of students is 50 </em></h2><h2><em>So the total area of the room is 9m²X50=450m² </em></h2><h2><em>Given the length of the room is 25m </em></h2><h2><em>so the Breadth is =450/25=18m </em></h2><h2><em> </em></h2><h2><em>each student requires 108m³ of space </em></h2><h2><em>so total volume of the room is 108X50=5400m³ </em></h2><h2><em>we know that : Volume=Area X h </em></h2><h2><em> so⇒5400=450Xh </em></h2><h2><em> ⇒h=5400/450= 12m</em></h2><h2><em /></h2><h2><em>HOPE IT HELPS (◕‿◕✿)</em></h2>

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2 years ago

Which graph represents this equation? <br> Equation: y = 2/3x^2 - 6x

vovangra [49]

Answer:

I'm afraid that none of them are, as the x-intercepts of the equation are 0 and 9 respectively

however, D. is the closest to it, and it would be the correct graph if there were parentheses around $x^{2}-6x$

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers