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3 years ago

Suppose another class

Mathematics

1 answer:

juin [17]3 years ago

5 0

Answer:

Step-by-step explanation:

22*0.2=4.4

22+4.4=26.4

when you round you will get 26

(you should probably ask your teacher which way to round.)

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How do I Round 53879 to the nearest hundred.

Elenna [48]

So,

53,879 rounded to the nearest hundred:

If the number in the tens place is greater than or equal to 5, then round up. If not, round down.

7 is greater than or equal to 5, so round up.

53,879 --> 53,900

3 0

3 years ago

Read 2 more answers

14) Rounded to the nearest whole number, the square

Katen [24]

Answer:

3934

Step-by-step explanation:

Round 15,479,652 and find square root.

4 0

2 years ago

boeing plans to increase its price for jetliners. with a selling price $201.5 million and a cost of $190.1 million, what was the

scoray [572]

so from 190.1 to 201.5 is 201.5 - 190.1 = 11.4.

now, if we take 190.1 as the 100%, what is 11.4 off of it in percentage?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 190.1&100\\ 11.4&x \end{array}\implies \cfrac{190.1}{11.4}=\cfrac{100}{x}\implies 190.1x=1140 \\\\\\ x=\cfrac{1140}{190.1}\implies x\approx 5.99684$

3 0

2 years ago

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Brent is saving for a computer that is $840. He will also need to pay a 6% sales tax. What is the total cost of the computer aft

Kobotan [32]

Answer:

$890.4

Step-by-step explanation:

840 + .06(840)

840 + 50.4

890.4

5 0

2 years ago

Read 2 more answers

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:

C. $(5-\frac{1}{2})^6$

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]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

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

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 $^6C_4$

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

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

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

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

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

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

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

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

Check the list of options for the expression on the left hand side

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

3 0

3 years ago