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2 years ago

10

A small Learning Center holds twenty-four less students than a large Learning Center. Together the Learning Centers hold seventy

-two students. How many students does each Learning Center hold?

1 answer:

olga2289 [7]2 years ago

6 0

Answer:

A small learning center holds 48 students

Step-by-step explanation:

A small Learning Center holds <em><u>twenty-four less</u></em> students <u><em>than a large Learning Center.</em></u> Together the <em><u>Learning Centers hold seventy-two</u></em> students. How many students does each Learning Center hold?

72-24=48

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Indicate the equation of the line, in standard form, that is the perpendicular bisector of the segment with endpoints (4, 1) and

jenyasd209 [6]

Equation of a line is $x+3y =-3$ .

<h3>What is a perpendicular bisector of the line segment?</h3>

A perpendicular bisector is a line that cuts a line segment connecting two points exactly in half at a 90 degree angle. To find the perpendicular bisector of two points, all you need to do is find their midpoint and negative reciprocal, and plug these answers into the equation for a line in slope-intercept form.

Given that,

Endpoints of the line segment are ( $x_{1},y_{1}$ ) = (4, 1) and ( $x_{2},y_{2}$ ) = (2, -5).

First find the midpoints of the given line segment.

M = $\left(\frac{x_{1}+x_{2} }{2},\frac{y_{1}+y_{2} }{2}\Right)$

= $\left(\frac{4+2 }{2},\frac{1-5 }{2}\Right)$

M = $(3,-2)$

Now, Find the slope of the line :

It is perpendicular to the line with (4,1) and (2,-5)

Slope between ( $x_{1},y_{1}$ ) and ( $x_{2},y_{2}$ ) = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

so,

the slope between (4,1) and (2,-5) = $\frac{-5-1 }{2-4 }$

= 3

perpendicular lines have slopes the multiply to get -1

3 times m=-1

m= $\frac{-1}{3}$

The equation of a line that has a slope of m and passes through the midpoints M(3,-2) is

$y-y_{1} =m(x-x_{1} )$

$y-(-2) =\frac{-1}{3} (x-3 )$

$(y+2) =\frac{-1}{3} (x-3 )$

if we want slope intercept form

$(y+2) =\frac{-1}{3} x+1$

$y= \frac{-1}{3} x-1$

If we want standard form

$\frac{1}{3} x+y = -1$

$x+3y =-3$

Hence, Equation of a line is $x+3y =-3$ .

To learn more about perpendicular bisector of the line segment from the given link:

brainly.com/question/4428422

#SPJ4

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8 months ago

Find the distance between Points C(4, 6) and D(−5, 6) without using a coordinate plane.

balu736 [363]

Answer:

-1, 0

Step-by-step explanation:

i think <33

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2 years ago

Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0

eduard

If you're using the app, try seeing this answer through your browser: brainly.com/question/2787701

_______________

$\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}$

Square both sides:

$\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}$

$\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\
\mathsf{17\,sin^2\,\theta=1}\\\\
\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}$

Since $\mathsf{sin\,\theta}$ is positive, you can discard the negative sign. So,

$\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}$

Substitute this value back into $\mathsf{(i)}$ to find $\mathsf{cos\,\theta:}$

$\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

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2 years ago

Greatest common factor for 40,10

horrorfan [7]

The greatest common factor of 40&10 is 10

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3 years ago

Read 2 more answers

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Bess [88]

27*.1= 2.70

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2 years ago