If you want this for $1 would u buy 10,000 of these ?add plz

Which expression could be substituted for y in the second equation to find the value of x?

EastWind [94]

value of x is $x=-\frac{29}{4}$ and value of y is $y=-\frac{9}{2}$

Step-by-step explanation:

We need to find Which expression could be substituted for y in the second equation to find the value of x

The given equations are:

$2x-y=-10\,\,\,(1)2x - 3y= -1\,\,\,\,(2)$

For putting value of y in second equation, we will find from equation 1

From equation 1:

$2x-y=-10\\-y=-10-2x\\y=10+2x$

Putting value of y i.e y=10+2x in equation (2) to find value of x

$2x-3y=-1\\2x-3(10+2x)=-1\\2x-30-6x=-1\\-4x=-1+30\\-4x=29\\x=\frac{29}{-4}\\x=-\frac{29}{4}$

So, now finding value of y by putting value of x= -29/4 in eq(1)

$2x-y=-10\\2(-\frac{29}{4})-y=-10\\-\frac{29}{2}-y=-10\\-y=-10+\frac{29}{2}\\ -y=\frac{-20+29}{2}\\ -y=\frac{9}{2}\\ y=-\frac{9}{2}$

So, value of x is $x=-\frac{29}{4}$ and value of y is $y=-\frac{9}{2}$

Keywords: System of equations

Learn more about System of equations at:

#learnwithBrainly

6 0

4 years ago

I need help with this!!! Thank you

ruslelena [56]

Answer:

4n^3-n^2-5n

Step-by-step explanation:

n(2n+4n2−5−3n)

=(n)(2n+4n2+−5+−3n)

=(n)(2n)+(n)(4n2)+(n)(−5)+(n)(−3n)

=2n2+4n3−5n−3n2

=4n3−n2−5n

6 0

3 years ago

When Jean was asked to say what the 3 in the fraction 2

JulijaS [17]

Fractions are quotients of two integers

A better way to say what the 3 in 2/3 represents is that, 3 is the denominator

<h3>How to determine what 3 represents</h3>

The fraction is given as: 2/3

A fraction is represented as: Numerator/ Denominator

By comparison;

Numerator = 2
Denominator = 3

Hence, a better way to say what the 3 in 2/3 represents is that, 3 is the denominator

Read more about fractions at:

brainly.com/question/11562149

7 0

2 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

3 years ago

Solve the literal equation for x. <br><br> k= nx - 3x

Amiraneli [1.4K]

answer:X=k/n-3

explain:move all terms to the left side and set equal to zero. then set each factor equ to zero!

I hope this helps

7 0

3 years ago

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