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3 years ago

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Evening tickets to a play are $24.50 each. Tickets for the afternoon show are $19 each. Janice wants to buy 6 tickets. Arrange t

erms from the options below to construct the expression Janice would use to determine how much less she would spend if she chooses an afternoon show instead of an evening show. (You do not need to use all of the terms offered as options. Use any term only once.)

1 answer:

kolbaska11 [484]3 years ago

3 0

Step-by-step explanation:

24.50x

x= 6

147

19x

x=6

114

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lbvjy [14]

Given:

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$(2x+3y)(2x-3y)=4x^2+6y^2$

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Solution:

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Taking left-hand side, we get

$L.H.S.=(2x+3y)(2x-3y)$

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$L.H.S.=(2)^2(x)^2-(3)^2(y)^2$ $[\because (ab)^x=a^xb^x]$

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It is not equal to right-hand side $4x^2+6y^2$ . In the right hand side, there must be a negative sign instead of positive sign.

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3 years ago

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Step-by-step explanation:

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yarga [219]

Because this is a fourth degree that will be tricky to factor as it is, we will do a u substitution. Let $x^2=u$ . We can now rewrite that polynomial in terms of u: $u^2-7u-8=0$ . Filling into the quadratic formula we have $u=\frac{7+/-\sqrt{49-4(1)(-8)}}{2}$ . Simplifying down $u=\frac{7+/-\sqrt{81}}{2}$ and $u=\frac{7+9}{2}$ or $u=\frac{7-9}{2}$ . That means that u = 8 or u = -1. But don't forget that we let $u=x^2$ , so we have to put x-squared back in for u. That gives us $x^2=8$ which simplifies down to $+/-2\sqrt{2}$ . That also gives us $x^2=-1$ . When we take the square root of -1, we have to use the fact that -1 = i^2, so we sub that in to get $x=+/-i$ . All in all, your solutions are as follows: $2\sqrt{2},-2\sqrt{2},i,-i$ which is choice b.

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