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3 years ago

Solve the linear equations 4x-3y=16 and x+y=-3

Mathematics

1 answer:

alexgriva [62]3 years ago

7 0

4x - 3y = 16

x + y = -3

Subtract y from both sides of the 2nd equation.

x = -3 - y

Now that we have a value of x, we can plug it into the first equation to solve for y.

4(-3 - y) - 3y = 16

Distributive property.

-12 - 4y - 3y = 16

Combine like terms.

-12 - 7y = 16

Add 12 to both sides.

-7y = 28

Divide both sides by -7.

y = -4

The value of y is -4, and now we can solve for the value of x.

Plug this value into either equation.

x + (-4) = -3

Add 4 to both sides.

x = 1

The value of x is 1.

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Determine whether the solution of -2x = -15 is<br> Greater than or less than -15

brilliants [131]

Positive numbers are always greater than -15. Therefore, it's a greater than.

-2x = - 15

Dividing both sides by -2, we get

x = 7.5

Negative cancels on both sides, so we get a positive equation.

Positive numbers are always greater than -15. Therefore, it's a greater than.

What is greater than or less than?

Greater than and less than are the comparison symbols.
When the number is bigger or smaller than the other, then greater than and less than symbols are used.
If the number is greater than the other, the greater than (>) symbol is used.
If the number is lesser than the other, the less than the (<) symbol is used.

To learn more about greater than and less than, visit: brainly.com/question/15746367

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1 year ago

Let the region R be the area enclosed by the function f(x)=ln(x) and g(x)= 1/2x-2. Find the volume of the solid generated when t

Neporo4naja [7]

Answer:

$V=61.66$

Step-by-step explanation:

This problem can be solved by using the expression for the Volume of a solid with the washer method

$V=\pi \int \limit_a^b[R(x)^2-r(x)^2]dx$

where R and r are the functions f and g respectively (f for the upper bound of the region and r for the lower bound).

Before we have to compute the limits of the integral. We can do that by taking f=g, that is

$f(x)=g(x)\\ln(x)=\frac{1}{2}x-2$

there are two point of intersection (that have been calculated with a software program as Wolfram alpha, because there is no way to solve analiticaly)

x1=0.14

x2=8.21

and because the revolution is around y=-5 we have

$R=ln(x)-(-5)\\r=\frac{1}{2}x-2-(-5)\\$

and by replacing in the integral we have

$V=\pi \int \limit_{x1}^{x2}[(lnx+5)^2-(\frac{1}{2}x+3)^2]dx\\$

$V=\pi [28x+\frac{1}{x}+xln^2x-12xlnx-6lnx]$

and by evaluating in the limits we have

$V=61.66$

Hope this helps

regards

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3 years ago

Find the commission on a $590.00 sale if the commission is 15%.

wariber [46]

Answer:

590 * 0.15 = $88.5

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The area of a triangular sign is 81ft2 if the base of the sign measures 18 what is the height of the sign

bagirrra123 [75]

The answer would be 9. Hope that helped!

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Katie draws two chords in a circle in a certain way. She then draws in four segments between adjacent endpoints of the chords to

FromTheMoon [43]

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3 years ago