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Nitella [24]
2 years ago
14

A line that includes the point (-6, 10) has a slope of -7. What is its equation in point-slope form?

Mathematics
2 answers:
choli [55]2 years ago
8 0

Answer:

y - 10 = -7(x + 6)

Step-by-step explanation:

klemol [59]2 years ago
3 0
Y-10=-7(x+6)

Hope it helps!
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Jk=12 j lies at 4 where could k be located ?
NikAS [45]

Answer:

16,-8

Step-by-step explanation:

jk=12

j lies at 4

k lies at 4+12=16

or k lies at 4-12=-8

8 0
3 years ago
What would be the probability for two even numbers.?
frozen [14]
Sorry, I didn't read the question fully. I don't know how to do this. Please disregard my previous answer.
6 0
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I NEED HELP WILL GIVE YOU 100 POINTS!!!! IF SOLVED CORRECTLY ​
Dafna11 [192]

Answer:

The edge lengths are 3 for all of them. Radical cubed 27 is also 3.

Step-by-step explanation:

27/ 3 = 9.  The  square root of 9 is 3.  So 3 * 3 * 3 = 27. For the bottom answer, it is 3 since 3 * 3 * 3 = 27.

How simple is this answer. I don't understand your problem.

6 0
3 years ago
Graph the line with slope -1/2<br> passing through the point (4, -3).
serious [3.7K]

Answer:

see image

Step-by-step explanation:

5 0
2 years ago
What is the area of a triangle whose vertices are (4,0), (2,
yan [13]

Answer:

The correct option is;

b) 0 sq, units

Step-by-step explanation:

The vertices of the triangle are;

(4, 0), (2, 3), (8, -6)

The distance formula fr finding the length of a segment is given as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Where, (x₁, y₁) and (x₂, y₂) are the coordinates of the end points of the line

For the points (4, 0) and (2, 3) , we have;

√((3 - 0)² + (2 -4)²) = √13

Distance from (4, 0) to (2, 3) = √13

For the points (4, 0) and (8, -6) , we have;

√((-6 - 0)² + (8 -4)²) = √13 =

Distance from (4, 0) to (8, -6) = 2·√13

For the points (2, 3) and (8, -6) , we have;

√((-6 - 3)² + (8 -2)²) = 3·√13 =

Distance from (2, 3) to (8, -6) = 3·√13

Therefore, the perimeter of the triangle = 6·√13

The semi perimeter s = 3·√13

The area of the triangle,  A  = \sqrt{s\cdot \left (s-a  \right )\cdot \left (s-b  \right ) \cdot \left ( s-c  \right )}

Where;

a, b, and c are the length of the sides of the triangle;

A  = \sqrt{3\cdot \sqrt{3} \cdot \left (3\cdot \sqrt{3} -\sqrt{3}   \right )\cdot \left (3\cdot \sqrt{3} -2 \cdot \sqrt{3}   \right ) \cdot \left ( 3\cdot \sqrt{3} -3\cdot \sqrt{3}   \right )} = 0

Therefore, the area = 0 sq, units.

6 0
3 years ago
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