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Gwar [14]
3 years ago
13

Solve for y 2x+y=3 thank you.

Mathematics
2 answers:
xxMikexx [17]3 years ago
8 0
Y = -2x + 3
Hope that helps.
Alexxx [7]3 years ago
4 0
Hello there!

\boxed{    2*x+y-(3)=0 } \\ \\ \boxed{ Solve \ :   2x+y-3  = 0 } \\ \\ y=? \\ \\ \boxed{  y-intercept = -2x+3} \\ \\ y=  \ \ \ \ \ \ \ \ \ \ \  \ \ \ \Uparrow\ \  \ \ \  \ \ \ \ \ \Uparrow

I hope this helps you!
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Find dy /dx of x² - 2x + 5 - 6x
BaLLatris [955]

Answer:

Step-by-step explanation:

If y=x^2-2x+5-6x then

y=x^2-8x+5 by combining like terms. (I'm thinking you might have mistyped the problem because this representation is weird).

\frac{dy}{dx}=2x-8

6 0
3 years ago
Nancy wants to sell 54 CD'S to her 6 friends for 5.00 each. How much will Nancy receive from selling her CD's? If each friend bu
alekssr [168]

Answer:

she will make $270 and each friend buys 9 CD's

Step-by-step explanation:

To find the money she makes take the CDs and multiply it times the money each one costs. To find the amount each friend buys divide the cds by th amount of friends

8 0
3 years ago
What lowest terms 19/31
jonny [76]
The lowest terms for 19/31 would be 1
7 0
3 years ago
Find an equation in standard form for the hyperbola with vertices at (0, ±3) and foci at (0, ±7)
Nitella [24]

The equation of a hyperbola is:

(x – h)^2 / a^2 - (y – k)^2 / b^2 = 1

 

So what we have to do is to look for the values of the variables:

<span>For the given hyperbola : center (h, k) = (0, 0)
a = 3(distance from center to vertices)
a^2 = 9</span>

<span>
c = 7 (distance from center to vertices; given from the foci)
c^2 = 49</span>

 

<span>By the hypotenuse formula:
c^2 = a^2 + b^2
b^2 = c^2 - a^2 </span>

<span>b^2 =  49 – 9</span>

<span>b^2  = 40

</span>

Therefore the equation of the hyperbola is:

<span>(x^2 / 9) – (y^2 / 40) = 1</span>

5 0
3 years ago
Read 2 more answers
Show 2 different solutions to the task.
laila [671]

Answer with Step-by-step explanation:

1. We are given that an expression n^2+n

We have to prove that this expression is always is even for every integer.

There are two cases

1.n is odd integer

2.n is even integer

1.n is an odd positive integer

n square is also odd integer and n is odd .The sum of two odd integers is always even.

When is negative odd integer then n square is positive odd integer and n is negative odd integer.We know that difference of two odd integers is always even integer.Therefore, given expression is always even .

2.When n is even positive integer

Then n square is always positive even integer and n is positive integer .The sum of two even integers is always even.Hence, given expression is always even when n is even positive integer.

When n is negative even integer

n square is always positive even integer and n is even negative integer .The difference of two even integers is always even integer.

Hence, the given expression is always even for every integer.

2.By mathematical induction

Suppose n=1 then n= substituting in the given expression

1+1=2 =Even integer

Hence, it is true for n=1

Suppose it is true for n=k

then k^2+k is even integer

We shall prove that it is true for n=k+1

(k+1)^1+k+1

=k^1+2k+1+k+1

=k^2+k+2k+2

=Even +2(k+1)[/tex] because k^2+k is even

=Sum is even because sum even numbers is also even

Hence, the given expression is always even for every integer n.

3 0
3 years ago
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