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Alja [10]
2 years ago
15

A line which has an undefined slope is _____.

Mathematics
2 answers:
Len [333]2 years ago
8 0
I think it’s a correct me if I’m wrong
erastovalidia [21]2 years ago
5 0

Answer:

c, vertical

Step-by-step explanation:

blablablablablanlabla

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Write the equation of the line that is parallel to y = -2x - 9 and passes through the point (-2,-4).
Cloud [144]

Answer:  y=-2x-8

 

Step-by-step explanation:

Parallel lines need to have the same slopes but different y-intercept.

y=-2x -9 is parallel to y=-2x - 8

7 0
3 years ago
Simplify the expression. Write your answer using an improper fraction<br> with no spaces.
ryzh [129]

Answer: 59/3k+14

Step-by-step explanation:

4 0
2 years ago
I need help finding the answers someone plz help
Harlamova29_29 [7]

Happy New Year from MrBillDoesMath!

Answer:

Proof by ASA congruence postulate. See below

Discussion:

  Fact 1 :     angle A = angle T             (given)

  Fact 2:    The angles on both sides of point X are equal as vertical angles

                 are equal.

From these facts it follows that angle M = angle H (as all plane triangles have 180 degrees). Also  AM = TH  (given) so

   In the left triangle                          In the right triangle

  (angle M, side AM, angle A)  =      (angle H, side TH, angle T)


Hence the triangles have two congruent angles, and congruent sides included between the angles, so they are congruent by  ASA.


Thank you,

MrB

8 0
3 years ago
HELP QUICKLY PLEASE!!! for y = 3x - 2x^2 + 5x^3, show how to find the value for y, when x = 3. please show your work.
Brrunno [24]
All you need to do is plug in 3 wherever you see an x!
y = 3(3) - 2(3)^2 + 5(3)^3
y = 9 - 2(9) + 5(27)
y = 9 - 18 + 135
y = -9 + 135
y = 126
Hope this helps!
7 0
3 years ago
Read 2 more answers
Given the vectors A⃗ and B⃗ shown in the figure ((Figure 1) ), determine the magnitude of B⃗ −A⃗. A is 28 degrees above the posi
Vlad [161]

This problem is represented in the Figure below. So, we can find the components of each vector as follows:


\bullet \ cos(28^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{A_{x}}{44} \\ \\ \therefore A_{x}=44cos(28^{\circ})=38.85m \\ \\ \\ \bullet \ sin(28^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{A_{y}}{44} \\ \\ \therefore A_{y}=44sin(28^{\circ})=20.65m


\bullet \ cos(56^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{-B_{x}}{26.5} \\ \\ \therefore B_{x}=-26.5cos(56^{\circ})=-14.81m \\ \\ \\ \bullet \ sin(56^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{B_{y}}{26.5} \\ \\ \therefore B_{y}=26.5sin(56^{\circ})=21.97m


Therefore:

\vec{A}=(38.85, 20.65)m \\ \\ \vec{B}=(-14.81, 21.97)m


So:

\vec{B}-\vec{A}=(-14.81, 21.97)-(38.85, 20.65)=(-53.66,1.32)


Finally, the magnitude is:


\boxed{\left| \vec{B}-\vec{A}\right|=\sqrt{(-53.66)^2+(1.32)^2}=53.67m}

7 0
3 years ago
Read 2 more answers
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