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3 years ago

PLEASE HELP K12

Mathematics

2 answers:

BartSMP [9]3 years ago

7 0

Answer:

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have the points (12, 6) and (12, -9). Substitute:

$d=\sqrt{(12-12)^2+(-9-6)^2}=\sqrt{0^2+(-15)^2}=\sqrt{225}=15$

DerKrebs [107]3 years ago

4 0

$\text{Hey there!}$

$\text{The formula is:}\bf{\frac{y_2-y_1}{x_2-x_1}}$ $\text{ or you can say}\rightarrow$ $\frac{Rise}{Run}$

$\text{y}_2=-9\\\\\text{y}_1=6\\\\\text{x}_2=12\\\\\text{x}_1=12$

$\text{Now, that we know what our points are we could solve this equation}$

$\frac{-9-6}{12-12}$

$\text{-9 - 6 = -15}$

$\text{12 - 12 = 0}$

$\bf{= \ \frac{-15}{0}}$

$\boxed{\boxed{\bf{Answer: \ \frac{-15}{0}}}}\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{LoveYourselfFirst:)}$

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Monica [59]

Answer:

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3 years ago

An arc 28 cm is cut from a circle of radius 15 cm. Find the angle of the sector formed by this arc.

Nutka1998 [239]

Answer:

$\displaystyle \frac{28}{15}\; \text{radians}$ , which is approximately $107^{\circ}$ .

Step-by-step explanation:

In a given circle, the angle of a sector is proportional to the length of the corresponding arc:

$\displaystyle \frac{\text{angle of sector $1$}}{\text{angle of sector $2$}} = \frac{\text{length of arc of sector $1$}}{\text{length of arc of sector $2$}}$ .

For the circle in this question, $r = 15\; \rm cm$ , and the circumference would be:

$2\, \pi \, r = 30\, \pi\; \rm cm$ .

The full circle itself is like a sector with an angle of $2\, \pi$ , with the arc length equal to the circumference of the circle.

$\displaystyle \frac{\text{angle of sector}}{\text{angle of full circle}} = \frac{\text{length of arc of sector}}{\text{circumference of circle}}$ .

$\displaystyle \frac{\text{angle of sector}}{2\,\pi\; \rm rad} = \frac{28\; \rm cm}{30\, \pi\; \rm cm}$ .

Rearrange the equation to find the angle of the sector:

$\begin{aligned} & \text{angle of sector} \\ =\; & (2\,\pi\; \rm rad) \cdot \frac{28\; \rm cm}{30\, \pi\; \rm cm} \\ =\; & \frac{28}{15}\; \rm rad\end{aligned}$ .

In other words, the angle of this sector would be $\displaystyle \frac{28}{15}\; \rm rad$ . Multiply that measure in radians by $\displaystyle \frac{360^{\circ}}{2\,\pi\; \rm rad}$ to find the value of the angle measured in degrees:

$\begin{aligned} & \frac{28}{15}\; \rm rad \times \frac{360^{\circ}}{2\,\pi\; \rm rad} \approx 107^{\circ}\end{aligned}$ .

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3 years ago

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Dmitry_Shevchenko [17]

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3 years ago

Read 2 more answers

A certain test preparation course is designed to help students improve their scores on the USMLE exam. A mock exam is given at t

bogdanovich [222]

Answer:

The critical value is $T = 4.604$ .

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

Step-by-step explanation:

The first step to solve this question is finding the sample mean and sample standard deviation:

14,11,18,9,19

Sample mean is sum of all values divided by the number of values. Thus:

$\overline{x} = \frac{14+11+18+9+19}{5} = 14.2$

The sample standard deviation is the square root of the division of the sum of the subtractions squared of each value and the mean, and the number of values. Thus:

$s = \sqrt{\frac{(14-14.2)^2+(11-14.2)^2+(18-14.2)^2+(9-14.2)^2+(19-14.2)^2}{5}} = 3.868$

Confidence interval:

We have the standard deviation for the sample, and so we use the t-distribution to build the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 5 - 1 = 4

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 4 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 4.604.

The critical value is $T = 4.604$ .

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 4.604\frac{3.868}{\sqrt{5}} = 7.964$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 14.2 - 7.964 = 6.236.

The upper end of the interval is the sample mean added to M. So it is 14.2 + 7.964 = 22.164.

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

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3 years ago

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jasenka [17]

Answer:

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Step-by-step explanation:

Given statement is At least one gift in the bag is wrapped is true

The negation to the given statement is " At least one gift in the bag is not wrapped" - False

<u>To determine whether the statement is a negation of the given statement is True or False :</u>

<h3> At least one gift in the bag is not wrapped - False </h3><h3> Not every gift in the bag is wrapped-True </h3><h3>Every gift in the bag is wrapped-False </h3><h3>None of the gifts in the bag are wrapped-False</h3>

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3 years ago