Rounded 121580 to the nearest thousand

Vikentia [17]

Step-by-step explanation:

add up all of the numbers in the five groups then divide the numbers by the number of numbers you have for example if the sum of all of those numbers is let's a 25 and you have 5 numbers you'll divide 25 by 5 and your men will be 5:00 and if the sum of the numbers was 30 and you have six numbers you'll / 30by6 then your answer then the mean will be six

7 0

2 years ago

Read 2 more answers

Select the correct answer from each drop-down menu. Right triangle ABC is represented with the right angle at vertex B. Base BC

Svetradugi [14.3K]

By using trigonometric relations, we will see that:

AC = 15.6 in

AB = 8.4 in.

<h3>How to get the measures of the other two sides of the right triangle?</h3>

Here we have the right triangle where:

B = 90°

C = 40°

BC = 10 in.

Notice that is the adjacent cathetus to the angle C, then we can use the two relations:

sin(a) = (adjacent cathetus)/(hypotenuse).
tan(a) = (opposite cathetus)/(adjacent cathetus).

Where:

hypotenuse = AC
opposite cathetus = AB.

Then we will have:

sin(40°) = 10in/AC.

AC = 10in/sin(40°) = 15.6 in

tan(40°) = AB/10in

tan(40°)*10in = AB = 8.4 in.

So we can conclude that for the given right triangle we have:

AC = 15.6 in

AB = 8.4 in.

If you want to learn more about right triangles:

brainly.com/question/2217700

#SPJ1

6 0

1 year ago

Can anyone figure this out?

Verizon [17]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill$

$\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}$

now that we know how long each one is, let's plug those in Heron's Area formula.

$\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18$