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Norma-Jean [14]
3 years ago
5

Choose the correct slope of the line that passes through the points (1, −3) and (3, −5).

Mathematics
2 answers:
Lelu [443]3 years ago
7 0
I believe it is -1.
Sladkaya [172]3 years ago
5 0
Slope of a line passing though point (x1,y1) nd (x2,y2) is
(y2-y1)/(x2-x1)

we have
(1,-3) and (3,-5)
x1=1
y1=-3
x2=3
y2=-5

slope=(-5-(-3))/(3-1)=(-5+3)/(2)=-2/2=-1

slope=-1
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4x + 5y = 10
Vika [28.1K]

Answer:

For A = 32/5 and B = 8 the system of equations will have infinitely many solutions.

Step-by-step explanation:

Given equations are:

4x + 5y = 10

Ax + By = 16

The general form of linear equation in two variables is given by:

ax+by = c

Here a, b and c are constants and x,y are variables.

In the given equations, after comparing with the general form

a_1 = 4\\b_1 = 5 \\c_1 = 10\\a_2 = A\\b_2 =B\\c_2 = 16

"In order for a system of equations to have infinity many solutions,

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} "

Putting the values we get

\frac{4}{A} = \frac{5}{B} = \frac{10}{16}\\\frac{4}{A} = \frac{5}{B} = \frac{5}{8}\\Now\\\frac{4}{A} = \frac{5}{8}\\\frac{A}{4} = \frac{8}{5}\\A = \frac{8}{5} * 4\\A = \frac{32}{5}\\And\\\frac{5}{B} = \frac{5}{8}\\\frac{B}{5} = \frac{8}{5}\\B = 8

Hence,

For A = 32/5 and B = 8 the system of equations will have infinitely many solutions.

5 0
2 years ago
Compare simplifying before multiplying fractions with simplifying after multiplying the fractions
Kipish [7]
<span>We will start using a new way to indicate simplifying fractions. When a numerator or 
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5 0
3 years ago
Read 2 more answers
Create your own GCF word problem with 3 numbers
svet-max [94.6K]
Number set:
18, 36, 900
GCF:
9
5 0
3 years ago
How do i solve this problem
Vladimir79 [104]
Solve the following system:
{6 t - 5 s = -4 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{-2 r - 4 s - 4 t = -9 | (equation 3)

Swap equation 1 with equation 3:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Subtract 1/2 × (equation 1) from equation 2:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 1 by -1:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 2 by 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 4 s + 10 t = 1 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Swap equation 2 with equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r - 4 s + 10 t = 1 | (equation 3)
Subtract 4/5 × (equation 2) from equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+(26 t)/5 = 21/5 | (equation 3)

Multiply equation 3 by 5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+26 t = 21 | (equation 3)
Divide equation 3 by 26:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s+0 t = (-115)/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 2 by -5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 2) from equation 1:
{2 r + 0 s+4 t = 25/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 3) from equation 1:
{2 r+0 s+0 t = (-17)/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 1 by 2:
{r+0 s+0 t = (-17)/26 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
v0 r+0 s+t = 21/26 | (equation 3)
Collect results:Answer:  {r = -17/26
               {s = 23/13                        {t = 21/26
7 0
3 years ago
Please help I’m having trouble with this question!
mezya [45]

ok thanks for letting me know you need help

4 0
2 years ago
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