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aleksklad [387]

3 years ago

9

Solve the system of equations below algebraically 2x+3y=6 and -5x+2y=4

1 answer:

Charra [1.4K]3 years ago

7 0

Steps
I’m using elimination so I’m trying to get rid of y
To do that I multiply the first equation by 2 and the second by -3. Then solve
4x + 6x = 12
15x + -6x = -12
—————————
19x + 0 = 0
Divid by 19 to get x to one side. 0/19 is 0 so
X= 0
Plug that into an equation so 0 + 3y = 6
Divid by 3 and y = 2
Answer x = 0 y = 2

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Solve for Y: 12y+d=−19y+t

Elina [12.6K]

Answer:

y = t/31 - d/31

Step-by-step explanation:

Solve for y:

d + 12 y = t - 19 y

Subtract d - 19 y from both sides:

31 y = t - d

Divide both sides by 31:

Answer: y = t/31 - d/31

4 0

3 years ago

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Mr mistry has 3 grandchildren. He gives Nimisha £m. He gives Sunhil £80 less than Nimisha. He gives Akshay twice as much as Sunh

slamgirl [31]

Here are the variables.

M= nimisha money

S= Sunhil money

A= Akshay money

The expression is S= M-80
A= S*2
M=S+80

Hope this helped

7 0

3 years ago

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The expression $16x + $58.75 represents the total amount of money the academic club earned from washing x cars at a fundraiser.

umka21 [38]

Answer:

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Step-by-step explanation:

3 0

3 years ago

Determine the intercepts of the line 2x+5y=-6?

SVETLANKA909090 [29]

Answer:

The x intercept is (-3,0)

The y intercept is (0,-6/5)

Step-by-step explanation:

To find the x intercept, set y =0 and solve for x

2x+5y=-6

2x+0 = -6

Divide by 2

2x/2 = -6/2

x = -3

The x intercept is (-3,0)

To find the y intercept, set x =0 and solve for y

2x+5y=-6

0+5y = -6

Divide by 5

5y/5 = -6/5

y = -6/5

The y intercept is (0,-6/5)

7 0

3 years ago

I need this answer asap

padilas [110]

Answer:

$a=64,b=0$

Step-by-step explanation:

Given;

$z=-1-\sqrt{3}i$

$r=\sqrt{(-1)^2+(-\sqrt{3})^2} =2$

$\phi =\tan^{-1}(\frac{-\sqrt{3} }{-1})=\frac{\pi}{3}$

$\arg(z)=2\pi-\frac{\pi}{3} =\frac{5\pi}{3}$

Apply DeMoivre's Theorem;

$z^n=r^n(\cos(n \theta) + i\sin(n \theta))$

$\Rightarrow z^6=2^6(\cos(6\times \frac{5\pi}{3}) + i\sin(6\times \frac{5\pi}{3}))$

$\Rightarrow z^6=64(\cos(10\pi) + i\sin(10\pi))$

$\Rightarrow z^6=64(1+ 0i)$

$\Rightarrow z^6=64+ 0i$

$\therefore a=64,b=0$

7 0

3 years ago