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4 years ago

Solve 5x − 6y = −38 3x + 4y = 0

Mathematics

1 answer:

Leto [7]4 years ago

7 0

Answer:

x = -76 , y = 57

Step-by-step explanation:

5x - 6y = -38

3x + 4y = 0

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

use 4 to multiply equation 1

5x(4) - 6y(4) = - 38(4)

20x -24y = -152

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

use 6 to multiply equation 2

3x(6) + 4y(6) = 0(6)

18x + 24y = 0

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

20x - 24y = -152

18x + 24y = 0

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

2x = -152

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

divide 2 by both sides

x = -76

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to find y, slot in the of x to either equation 1 or 2

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

equation 2

3x + 4y = 0

3(-76) + 4y = 0

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

-228 + 4y = 0

4y = 228 - 0

4y = 228

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divide both sides by 4

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

y = 57

for this type of question, you can use

substitution method
elimination method
graphical method

i used substitution method

hope it helps

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4 years ago

1. Suppose D has a coordinate of 11 and DE = 5. What are the possible midpoints of DE?

Rasek [7]

Answer:

The possible midpoints are (5.5,8) and (5.5,3)

Step-by-step explanation:

Given

$D = 11$

$DE = 5$

Required

Determine the midpoint of DE

First, we need to determine the coordinates of E

If E is to the right of DE;

$E - D = 5$

$E - 11 = 5$

$E = 11 + 5$

$E = 16$

Hence, the coordinate is (11,16)

$Midpoint = \frac{1}{2} * (11,16)$

$Midpoint = (5.5,8)$

Else:

$D - E = 5$

$11 - E = 5$

$E = 11 - 5$

$E = 6$

Hence, the coordinate is (11,6)

$Midpoint = \frac{1}{2} * (11,6)$

$Midpoint = (5.5,3)$

The possible midpoints are (5.5,8) and (5.5,3)

6 0

3 years ago

The graph of f(x) = 24x^2 + 63x +34/ 4x + 3 has an oblique asymptote at y =

g100num [7]

Correct answer is C). 6X+11

Step-by-step explanation:

The given function f(x)= $\frac{24x^{2} +63x + 34 }{4x+3}$

On simplying the function f(x),

f(x)= $\frac{6x(4x+3) -18x +63x + 34 }{4x+3}$

f(x)= $\frac{6x(4x+3) +45x + 34}{4x+3}$

f(x)= $\frac{6x(4x+3) +11(4x+3)+x-33 + 34 }{4x+3}$

f(x)= $\frac{6x(4x+3) +11(4x+3)+x+1}{4x+3}$

f(x)= $(6x+11)+\frac{(x+1)}{4x+3}$

From above equation

"(6x+11)" is Oblique asymptote and $\frac{(x+1)}{4x+3}$ is vertical asymptote

Therefore, Correct answer is C). 6X+11

8 0

3 years ago

Read 2 more answers