All categories

3 years ago

An equivalent equation has been written by multiplying the second equation by 2.

Mathematics

1 answer:

sukhopar [10]3 years ago

5 0

Answer:

(-19/2, -5)

Step-by-step explanation:

Given the two equivalent equations

4x – 9y = 7 .... 1

–4x + 6y = 8 ....2

Adding both equations

-9y + 6y = 7+8

-3y = 15

y = -15/3

y = -5

Substitute y = -5 into 1

From 1; 4x - 9y = 7

4x - 9(-5) = 7

4x + 45 = 7

4x = 7 - 45

4x = -38

x = -38/4

x = -19/2

Hence the solution is (-19/2, -5)

You might be interested in

For what value of x is the rational expression below undefined?

Blizzard [7]

Answer: The expression is undefined for x=4 and x=5.

The expression is undefined for any x that makes the denominator 0. This leads to solving a quadratic equation:

$\frac{x+3}{x^2-9x+20}\\x^2-9x+20\neq 0\\x_{1,2}\neq\frac{9\pm\sqrt{9^2-80}}{2}=\frac{9\pm1}{2}\\x_1\neq4\\x_2\neq5$

4 0

4 years ago

Read 2 more answers

Ed intends to roll a six-sided number cube 100 times. What probability model can he use to predict whether or not each roll will

Stells [14]

The answer is Each roll has a 0.333 probability of being composite.

<em>(I just took the test, B was the answer)</em>

8 0

3 years ago

Read 2 more answers

write the equation of the line, with the given properties, in slope-intercept form. slope = -5, through (-4,4)

Daniel [21]

Answer: y=-5x-16

Step-by-step explanation:

Since we are given the slope and the point, we can plug in the point to solve for y-intercept.

4=-5(-4)+b [multiply]

4=20+b [subtract both sides by 20]

b=-16

Now, we know the equation is y=-5x-16.

3 0

4 years ago

Help me please..!!!!

wariber [46]

I think it is D but i am not too sure

3 0

3 years ago

A particle moves so that r(t) = ati + b sin atj. Show that the magnitude of the acceleration of the particle is proportional to

ValentinkaMS [17]

Answer:

Step-by-step explanation:

Given

Position of particle is $r(t)=at\hat{i}+b\sin (at)\hat{j}$

i.e. distance from x axis is $b\sin (at)---1$

Distance from y axis at

velocity is given by $v=\frac{\mathrm{d} r}{\mathrm{d} t}$

$v=a\hat{i}+ba\cos (at)\hat{j}$

Similarly acceleration is given by

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a=0\hat{i}-a^2b\sin (at)\hat{j}$

Magnitude of acceleration is $=\sqrt{(-a^2b\sin (at))^2}$

$=a^2b\sin (at)----2$

From 1 and 2 we can see that

Magnitude of acceleration is proportional to distance from x axis

$a\propto distance\ from\ x-axis$

5 0

3 years ago

Read 2 more answers