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drek231 [11]
3 years ago
11

Find the point, M, that divides segment AB into a ratio of 5:6 if A is at (0, 22) and B is at (11, 0). A) (5, 12) B) (5, 11) C)

(6, 12) D) (6, 11)
Mathematics
2 answers:
amid [387]3 years ago
6 0
\bf \left. \qquad  \right.\textit{internal division of a line segment}
\\\\\\
A(0,22)\qquad B(11,0)\qquad
\qquad 5:6\quad \textit{from A to B}
\\\\\\
\cfrac{AM}{MB} = \cfrac{5}{6}\implies \cfrac{A}{B} = \cfrac{5}{6}\implies 6A=5B\implies 6(0,22)=5(11,0)\\\\
-------------------------------\\\\

\bf { M=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}\\\\
-------------------------------\\\\
M=\left(\cfrac{(6\cdot 0)+(5\cdot 11)}{5+6}\quad ,\quad \cfrac{(6\cdot 22)+(5\cdot 0)}{5+6}\right)
\\\\\\
M=\left( \cfrac{55}{11}~~,~~\cfrac{132}{11} \right)

and surely you know what that is.
lozanna [386]3 years ago
4 0

Answer:

Answer is (5,12) if anyone wanted to know

Step-by-step explanation:


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Use integration by parts to find the integrals in Exercise.<br> ∫(x+6)ex dx.
skelet666 [1.2K]

Answer:

xe^x+5e^x

Step-by-step explanation:

We have given integral \int (x+6)e^xdx

\int (xe^x+6e^x)dx

\int (I_1+I_2)dx , here I_1=xe^x and I_2=6e^x

Now first integrate I_1

So I_1=\int xe^x

Integrating by part

I_1=x\int e^x-\int e^x\frac{d}{dx}x

I_1=x e^x- e^x

I_2=6\int e^x=6e^x

So I=I_1+I_2=xe^x-e^x+6e^x=xe^x+5e^x

5 0
3 years ago
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lozanna [386]

Answer:

c = 60

Step-by-step explanation:

A triangle = 180 degrees

180-(15+105)=c

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5 0
3 years ago
Please help will mark Brainly
Troyanec [42]

Answer:

not

Step-by-step explanation:


6 0
4 years ago
Please help me answer one of these!
babunello [35]

This seems to be referring to a particular construction of the perpendicular bisector of a segment which is not shown. Typically we set our compass needle on one endpoint of the segment and compass pencil on the other and draw the circle, and then swap endpoints and draw the other circle, then the line through the intersections of the circles is the perpendicular bisector.


There aren't any parallel lines involved in the above described construction, so I'll skip the first one.


2. Why do the circles have to be congruent ...


The perpendicular bisector is the set of points equidistant from the two endpoints of the segment. Constructing two circles of the same radius, centered on each endpoint, guarantees that the places they meet will be the same distance from both endpoints. If the radii were different the meets wouldn't be equidistant from the endpoints so wouldn't be on the perpendicular bisector.


3. ... circles of different sizes ...


[We just answered that. Let's do it again.]


Let's say we have a circle centered on each endpoint with different radii. Any point where the two circles meet will then be a different distance from one endpoint of the segment than from the other. Since the perpendicular bisector is the points that are the same distance from each endpoint, the intersection of circles with different radii isn't on it.


4. ... construct the perpendicular bisector ... a different way?


Maybe what I first described is different; there are no parallel lines.



8 0
4 years ago
Item 7
Katyanochek1 [597]

i am sorry i was gonna anwer but i keep getting it wrong mys elf

4 0
3 years ago
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