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1 year ago

Look at the figure. Which step should be taken next to construct a line through point P perpendicular to BA?

Mathematics

1 answer:

alexira [117]1 year ago

8 0

ANSWER:

C. Place the compass on point A. Open the compass to a point between point P and point B.

EXPLANATION:

A perpendicular is a line that would be at a right angle to line BA.

The next step is to chose a radius that is greater than PB or PA so as to construct the bisector. And this can be done by placing the compass on point A, and open the compass to a point between point P and point B.

Use this radius to draw an arc above and below the line, and repeat the same using B as the center with the same radius. This would form two intersecting arcs above and below line BA. Join the point of intersection of the arcs by a straight line through P. This is the bisector of line BA through point P.

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A sumo wrestling ring is circular and has a circumference of 4.6\pi \text{ meters}4.6π meters4, point, 6, pi, start text, space,

frez [133]

Answer:

The area of the sumo wrestling ring is $5.29 \pi$

Step-by-step explanation:

The circumference of the circular sumo wrestling ring is $4.6\pi$ , that means its radius $r$ is:

$2\pi r=4.6\pi$

$r=\frac{4.6}{2} =\boxed{2.3\:meters.}$

Now once we have the radius $r$ of the sumo wrestling ring we can find its area $A$ by the following formula:

$A=\pi r^2$

Putting in the value of $r=2.3\:meters$ we get:

$A=\pi (2.3m)^2=\boxed{5.29\pi\:\:m^2}$

Therefore the area of the sumo wrestling ring is ${5.29\pi\:\:m^2$

3 0

1 year ago

Read 2 more answers

There are 5 Snickers, 10 Baby Ruths, 13 Milky Ways, 12 Twixs and17 Almond Joys in a bowl of candy. You reach into the bowl and r

agasfer [191]

Answer: 0.2415

Step-by-step explanation:

Given : There are 5 Snickers, 10 Baby Ruths, 13 Milky Ways, 12 Twixs and17 Almond Joys in a bowl of candy.

Total candy bars : $5+10+13+12+17=57$

Probability of getting a Milky Way bar= $p=\dfrac{\text{No. of Milky bars}}{\text{Total candy bars}}$

$\Rightarrrow\ p=\dfrac{13}{57}\approx0.23$

Using Binomial distribution , the probability of getting success in x trials is given by:-

$P(x)=^nC_xp^x(1-p)^{n-x}$ , where p uis probability of success in each trial and n is sample size.

If you randomly select n= 5 candy bars, then the probability you select exactly 2 Milky Way bars Will be :_

$P(2)=^5C_2(0.23)^2(1-0.23)^{3}\\\\=\dfrac{5!}{2!(5-2)!}(0.23)^2(0.77)^3\\\\=0.241505957\approx0.2415$

Hence, the probability you select exactly 2 Milky Way bars = 0.2415

8 0

2 years ago

Which model represents the factors of x2 + 9x + 8?

Tomtit [17]

Answer: the answer is D

Step-by-step explanation:

5 0

1 year ago

Read 2 more answers

Consider the probability distribution of x, where x is the number of job applications completed by a college senior through the

Oksi-84 [34.3K]

The expected number out of 1000 selected college seniors that completed 1 job application through the career centers is given by 0.011 x 1000 = 11 which is close to 14.

The expected number out of 1000 selected college seniors that completed 2 job application through the career centers is given by 0.115 x 1000 = 115 which is far away from 15.

The expected number out of 1000 selected college seniors that completed 3 job application through the career centers is given by 0.123 x 1000 = 123 which is close to 130.

Therefore, the result that would be suprising is "15 seniors completed 2 job applications through the career center."

3 0

2 years ago

Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold: $\bullet$ $f(x)$ is divisible by $x^3$. $\bullet

frosja888 [35]

There seems to be one character missing. But I gather that f(x) needs to satisfy

• $x^3$ divides $f(x)$

• $(x-1)^3$ divides $f(x)^2$

I'll also assume f(x) is monic, meaning the coefficient of the leading term is 1, or

$f(x) = x^5 + \cdots$

Since $x^3$ divides $f(x)$ , and

$f(x) = x^3 p(x)$

where $p(x)$ is degree-2, and we can write it as

$f(x)=x^3 (x^2+ax+b)$

Now, we have

$f(x)^2 = \left(x^3p(x)\right)^2 = x^6 p(x)^2$

so if $(x - 1)^3$ divides $f(x)^2$ , then $p(x)$ is degree-2, so $p(x)^2$ is degree-4, and we can write

$p(x)^2 = (x-1)^3 q(x)$

where $q(x)$ is degree-1.

Expanding the left side gives

$p(x)^2 = x^4 + 2ax^3 + (a^2+2b)x^2 + 2abx + b^2$

and dividing by $(x-1)^3$ leaves no remainder. If we actually compute the quotient, we wind up with

$\dfrac{p(x)^2}{(x-1)^3} = \underbrace{x + 2a + 3}_{q(x)} + \dfrac{(a^2+6a+2b+6)x^2 + (2ab-6a-8)x +2a+b^2+3}{(x-1)^3}$

If the remainder is supposed to be zero, then

$\begin{cases}a^2+6a+2b+6 = 0 \\ 2ab-6a-8 = 0 \\ 2a+b^2+3 = 0\end{cases}$

Adding these equations together and grouping terms, we get

$(a^2+2ab+b^2) + (2a+2b) + (6-8+3) = 0 \\\\ (a+b)^2 + 2(a+b) + 1 = 0 \\\\ (a+b+1)^2 = 0 \implies a+b = -1$

Then $b=-1-a$ , and you can solve for a and b by substituting this into any of the three equations above. For instance,

$2a+(-1-a)^2 + 3 = 0 \\\\ a^2 + 4a + 4 = 0 \\\\ (a+2)^2 = 0 \implies a=-2 \implies b=1$

So, we end up with

$p(x) = x^2 - 2x + 1 \\\\ \implies f(x) = x^3 (x^2 - 2x + 1) = \boxed{x^5-2x^4+x^3}$

6 0

1 year ago