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2 years ago

Please help ill give brainliest thanks and all 5 stars:((((((

Mathematics

2 answers:

igomit [66]2 years ago

7 0

Answer:

Step-by-step explanation:

Okay, so you can see x= 26, so you half it. You do this because it says 1/2. Then it says plus 7, 7 plus 13 equals 20. Therefore y= 20

pashok25 [27]2 years ago

5 0

The answer would be 20
Because if x is 26 then your equation would now be y=1/2(26)+7
So no you would multiply 1/2 times 26 which gives you 13 okay so now your equation should be y=13+7 and now you just add 13+7 which gives you 20 and there you go

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A rectangular auditorium seats 1564 people. the number of seats in each row exceeds the number of rows by 1212. find the number

yanalaym [24]

<span>We can safely assume that 1212 is a misprint and the number of seats in a row exceeds the number of rows by 12. Let r = # of rows and s = # of seats in a row. Then, the total # of seats is T = r x s = r x ( r + 12), since s is 12 more than the # of rows. Then r x (r + 12) = 1564 or r**2 + 12*r - 1564 = 0, which is a quadratic equation. The general solution of a quadratic equation is: x = (-b +or- square-root( b**2 - 4ac))/2a In our case, a = 1, b = +12 and c = -1564, so x = (-12 +or- square-root( 12*12 - 4*1*(-1564) ) ) / 2*1 = (-12 +or- square-root( 144 + 6256 ) ) / 2 = (-12 +or- square-root( 6400 ) ) / 2 = (-12 +or- 80) / 2 = 34 or - 46 We ignore -46 since negative rows are not possible, and have: rows = 34 and seats per row = 34 + 12 = 46 as a check 34 x 46 = 1564 = total seats</span>

4 0

3 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

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