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3 years ago

9

Angles round a point will always add up to ________⁰

1 answer:

WINSTONCH [101]3 years ago

3 0

360 - because a right angle is 90° and there are 4 right angles in a circle , 90 x 4 = 360

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Can y’all help me please

nikdorinn [45]

Answer:

$4a^2b^2(ab^3-4a^3+3b)$

This is the answer,the option C

Step-by-step explanation:

8 0

3 years ago

Find the interquartile range of the data.

likoan [24]

The answer is 14. Here is how I work it out.

First we are going have to identify quartile 1 and quartile 3.

So after putting the numbers in order from least to greatest mark the number with a half way point.

This is optional but it will help us spot the sections better.

31,33,35,41,43,|46,48,49,49,50

Next put parentheses around the remaining groups of numbers.

(31,33,35,41,43,)|(46,48,49,49,50)

For the next step we have to find the median of each group.

(31,33,35,41,43,)|(46,48,49,49,50)

The median of each group, are called the quartiles, the median of the lower half is quartile 1, and the median of the upper half is quartile 3.

Now subtract quartile 1 from quartile 3.

49 – 35 = 14

So the interquartile range of this data set is 14.

4 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

Negative 9 plus 2 over 3 plus open parentheses 1 plus 1 close parenthesis squared

Alik [6]

Answer:

⁻ $\frac{19}{3}$

Step-by-step explanation:

⁻ $9 + \frac{2}{3} (1+1)^{2}$

⁻ $9 + \frac{2}{3}$ × $2^{2}$

⁻ $9 + \frac{2}{3}$ × $4$

⁻ $9$ $+$ $\frac{2 x 4}{3}$

⁻ $9 + \frac{8}{3}$

⁻ $\frac{19}{3}$

8 0

4 years ago

Read 2 more answers

Functions f(x) and g(x) are shown below:

hoa [83]

f(x) is a multiple of the cosine function. g(x) is a multiple of the sine function. Both the cosine function and the sine function have a maximum value of 1, so the magnitude of the multiple will determine which function has the largest maximum value.

The multiple for f is 2; the multiple for g is 3, which is greater than 2. Hence g(x) has a larger maximum value than f(x).

6 0

3 years ago