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faltersainse [42]
3 years ago
7

Find the slope of the line (-4, -1) (1, 4)

Mathematics
2 answers:
tatyana61 [14]3 years ago
8 0
The slope of the line would be 5/5 but simplified to 1

timofeeve [1]3 years ago
7 0

(y - yo) = m.(x - xo)

We have two points (-4, -1) and (1, 4)

(-1 - 4) = m.(-4 - 1)

(-5) = m.(-5)

m = (-5)/(-5)

m = 1

So the slope is 1

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Estimate the measure of
Lelu [443]

Answer:

148°

Step-by-step explanation:

it is very close to the 150 line and the option closest to 150 is 148

8 0
3 years ago
Read 2 more answers
Which of the following sets are subspaces of R3 ?
Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

\to A={(x,y,z)|3x+8y-5z=2} \\\\\to  for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                        =3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)

The set A is not part of the subspace R^3

for point B:

\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon  B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                             =-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0

\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon  B

The set B is part of the subspace R^3

for point C: \to C={(x,y,z)|x

In this, the scalar multiplication can't behold

\to for (-2,-1,2) \varepsilon  C

\to -1(-2,-1,2)= (2,1,-1) ∉ C

this inequality is not hold

The set C is not a part of the subspace R^3

for point D:

\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)

The scalar multiplication s is not to hold

\to for (-4, 1,2)\varepsilon  D\\\\\to  -1(-4,1,2) = (4,-1,-2) ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace R^3

For point E:

\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to  a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\

The  x_1, x_2 is the arbitrary, in which ax_1+bx_2is arbitrary  

\to a(x_1,0,0)+b(x_2,0,0) \varepsilon  E

The set E is the part of the subspace R^3

For point F:

\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon  F \\\\\to  a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))

The x_1, x_2 arbitrary so, they have ax_1+bx_2 as the arbitrary \to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F

The set F is the subspace of R^3

5 0
3 years ago
Two angles are supplementary the first angle measures 40 degrees what is the second angle
nydimaria [60]
The second angle would be 140⁰. When two angles are supplementary, they add up to 180⁰.
180⁰-40⁰=140⁰
4 0
3 years ago
Which pair of numbers is relatively prime? A. 42 and 77 B. 34 and 55 C. 45 and 102 D. 99 and 123
leonid [27]

Two pair of numbers are said to be relatively prime if there is no integer greater than 1, that divides them both.

Consider the given pair of the numbers to identify which pair is relatively prime.

1. Consider 42 and 77

7 is the number which divides 42 and 77 both. Therefore, they are not relatively prime.

2. Consider 34 and 55

Since, there is no number greater than 1, which divides both the numbers. So, they are relatively prime numbers.

3. Consider 45 and 102

3 is the number divides 45 and 102 both. Therefore, they are not relatively prime.

4.  Consider 99 and 123

3 is the number divides 99 and 123 both. Therefore, they are not relatively prime.

Therefore, Option B is the correct answer.

8 0
3 years ago
Explaining your answer can either include a counterexample, a diagram or a written explanation to support your answer.
anzhelika [568]

Answer:

sorry I didn't understanding this

6 0
3 years ago
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