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3 years ago

Find the slope for (3,4) and (3,-4)

Mathematics

1 answer:

saveliy_v [14]3 years ago

7 0

Answer:

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The problem:

Find the slope for the line going through (3,4) and (3,-4).

Step-by-step explanation:

Line up points vertically and subtract.

Then put 2nd difference over 1st.

( 3 , 4)

-(3 , -4)

------------

0 8

So the slope would have been 8/0 but this is undefined.

So the slope is undefined.

Also notice the x's are the same and the y's are difference so this is a vertical line. There is only rise in a vertical line and no run. Recall, slope is rise/run. You cannot divide by 0 so this is why we say the slope is undefined when the x's are always the same no matter the y.

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Add 60 5/9 54 1/3 59 1/3 56 1/6 _____ Thank You!

kykrilka [37]

Answer:

230 5/9

Step-by-step explanation:

after adding tou would get 230 5/9

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3 years ago

Look at the figure, DFGME. Find the length of EM.

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EM is 20 my friend aloha

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3 years ago

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Classify each pair of angles listed below as adjacent or vertical.

Roman55 [17]

Given:

Intersecting lines DA and CE.

To find:

Each pair of adjacent angles and vertical angles.

Solution:

Adjacent angles are in the same straight line.

Pair of adjacent angles:

(1) ∠EBD and ∠DBC

(2) ∠DBC and ∠CBA

(3) ∠CBA and ∠ABE

(4) ∠ABE and ∠EBD

Vertical angles are opposite angles in the same vertex.

Pair of vertical angles:

(1) ∠EBD and ∠CBA

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3 years ago

Point B has coordinates (1,2). The x-coordinate of point A is -8. The distance between point A and point B is 15 units. What

Xelga [282]

Answer:

The possible coordinates of point A are $A_{1} (x,y) = (-8, 14)$ and $A_{2} (x,y) = (-8, -10)$ , respectively.

Step-by-step explanation:

From Analytical Geometry, we have the Equation of the Distance of a Line Segment between two points:

$l_{AB} = \sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2}}$ (1)

Where:

$l_{AB}$ - Length of the line segment AB.

$x_{A}, x_{B}$ - x-coordinates of points A and B.

$y_{A}, y_{B}$ - y-coordinates of points A and B.

If we know that $l_{AB} = 15$ , $x_{A} = -8$ , $x_{B} = 1$ and $y_{B} = 2$ , then the possible coordinates of point A is:

$\sqrt{(1+8)^{2}+(2-y_{A})^{2}} = 15$

$81 + (2-y_{A})^{2} = 225$

$(2-y_{A})^{2} = 144$

$2-y_{A} = \pm 12$

There are two possible solutions:

1) $2-y_{A} = -12$

$y_{A} = 14$

2) $2 - y_{A} = 12$

$y_{A} = -10$

The possible coordinates of point A are $A_{1} (x,y) = (-8, 14)$ and $A_{2} (x,y) = (-8, -10)$ , respectively.

8 0

3 years ago

Find the area of the following shape.

ad-work [718]

Answer:

57 units^2

Step-by-step explanation:

First find the area of the triangle on the left

ABC

It has a base AC which is 9 units and a height of 3 units

A = 1/2 bh = 1/2 ( 9) *3 = 27/2 = 13.5

Then find the area of the triangle on the right

It has a base AC which is 6 units and a height of 1 units

A = 1/2 bh = 1/2 ( 6) *1 = 3

Then find the area of the triangle on the top

It has a base AC which is 3 units and a height of 3 units

A = 1/2 bh = 1/2 ( 3) *3 = 9/2 = 4.5

Then find the area of the rectangular region

A = lw = 6*6 = 36

Add them together

13.5+3+4.5+36 =57 units^2

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3 years ago

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