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Aleks04 [339]
4 years ago
8

In the trapezoid ABCD ( AB ∥ CD ) point M∈ AD , so that AM:MD=3:5. Line l ∥ AB and going trough point M intersects diagonal AC a

nd leg BC at points P and N respectively. Find: BC:BN
Mathematics
1 answer:
Ann [662]4 years ago
5 0

Answer:

BC : BN = 8 : 3

Step-by-step explanation:

Parallel lines divide any transversal into segments with the same ratio. That means ...

... BN : NC = AM : MD = 3 : 5

Since BC = BN + NC, the desired ratio is ...

... BC : BN = (BN+NC) : BN = (3+5) : 3

... BC : BN = 8 : 3

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A pond 85 feet across; 1 inch = 4 feet
enot [183]
Well... all u have to do is 85 divided by 4 which equals 21.25..... unless there is another part of the question!!
3 0
4 years ago
For every 14 ice cream cones you buy, you get 2 for free.how many cones will you get if you buy 56 cones
lianna [129]
Divide 56/14=4. That means that there are 4 14ns since you will get 2 for every 14 cones. Thay means that you will get 8 cones for free.
6 0
3 years ago
Graph and label the image of the figure below after a dilation by a factor of 1/2.
neonofarm [45]
Answer:

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

see graph below

Explanation:

Given:

The image of a quadrilateral on a coordinate plane

To find:

The coordinates of the new image after dilation of 1/2 have been applied to the original image.

Then graph the coordinates

First, we need to state the coordinates of the original image:

M = (3, -2)

F = (4, -2)

L = (1, -5)

W = (5, -5)

Next, we will apply a scale factor of 1/2:

\begin{gathered} Dilation\text{ rule:} \\ (x,\text{ y\rparen}\rightarrow(kx,\text{ ky\rparen} \\ where\text{ k = scale factor} \\  \\ scale\text{ factor = 1/2} \\ M^{\prime}\text{ = \lparen}\frac{1}{2}(3),\text{ }\frac{1}{2}(-2)) \\ M^{\prime}\text{ = \lparen}\frac{3}{2},\text{ -1\rparen} \\  \\ F\text{ = \lparen}\frac{1}{2}(4),\text{ }\frac{1}{2}(-2)) \\ F^{\prime}\text{ = \lparen2, -1\rparen} \end{gathered}\begin{gathered} L\text{ = \lparen}\frac{1}{2}(1),\text{ }\frac{1}{2}(-5)) \\ L^{\prime}\text{ = \lparen}\frac{1}{2},\text{ }\frac{-5}{2}) \\  \\ W\text{ = \lparen}\frac{1}{2}(5),\text{ }\frac{1}{2}(-5)) \\ W^{\prime}\text{ = \lparen}\frac{5}{2},\text{ }\frac{-5}{2}) \end{gathered}

The new coordinates:

M' (3/2, -1), F' (2, -1), L' (1/2, -5/2), W' (5/2, -5/2)

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

Plotting the coordinates:

3 0
1 year ago
Given g(x) = :-2 +4, solve for a when g(x) = 0.
evablogger [386]

Answer:

-7/4

Step-by-step explanation:

You are looking for the composite g(f(2)). The simplest way to solve this is to evaluate f(2) and enter the solution in to your g function.

g(f(2))=g(-(2)^2-2(2)+4)=g(-4-4+4)=g(-4)

g(-4)=4/(-4(-4)-2)=4/(16-2)=4/14=2/7

Therfor, g(f(2))=2/7  **I'm assuming the -4x-2 is all in the denominator of the g(x) function. If -2 is not in the denominator you would have

g(f(2))=4/(-4(-4)) -2=4/16 -2=1/4 -2=1/4-8/4= -7/4

8 0
3 years ago
73.18 minus 5 .23 multiply 9.34​
maks197457 [2]

Answer: 634.653

I don't know if this is right or not, but i hope this helps

4 0
4 years ago
Read 2 more answers
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