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vampirchik [111]
4 years ago
11

PLEASE HELP.

Mathematics
1 answer:
pentagon [3]4 years ago
3 0

Let x represent the number of shirts. Let y represent the number of pens.

If shirts are on sale for $11.99 each, then x shirts cost $11.99x.

If pants are on sale for $12.99 each, then y pants cost $12.99y.

The total cost is $(11.99x+12.99y).

Sarah can spend up to $65. Then an inequality that represents this situation is

11.99x+12.99y≤65 (this inequality holds when Sarah can spend $65 too)

or

11.99x+12.99y<65 (this inequality holds when Sarah can spend less than $65).

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(2t-3)^3, can you help me?, please?
grandymaker [24]

Answer:

8t^3-36t^2+54t-27

Step-by-step explanation:

(2t-3)^3

(2t-3)(2t-3)(2t-3)

Multiply the first two terms

(4t^2 -6t -6t+9)(2t-3)

Combine like terms

( 4t^2 -12t +9) ( 2t-3)

Multiply

2t*( 4t^2 -12t +9) -3 ( 4t^2 -12t +9)

Distribute

8t^3 -24t^2 +18t-12t^2 +36t-27

Combine like terms

8t^3-36t^2+54t-27

6 0
3 years ago
Read 2 more answers
So uhm idk this can some1 help
lys-0071 [83]

Answer:

Negative

Step-by-step explanation:

Negative-negative is always negative, and -1/3-4/5 is negative too. the second one, however, is positive.

8 0
4 years ago
Read 2 more answers
Which is an example of phototropism?
Aleks04 [339]

Answer:

i think it is A

Step-by-step explanation:

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4 0
4 years ago
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Factor completely<br> 486+108x+6x^2<br><br> Thank you!
viktelen [127]

Answer:

6(x + 9) (x + 9)

or

6 (x+9)^{2}

Step-by-step explanation:

468 + 108x + 6x^{2} \\6x^{2} +108x + 468

Greatest common factor of 6, 108, and 486:

6 x 1 = 6

6 x 18 = 108

6 x 81 = 486

GCF: 6

so:

6(x^{2} +18x +81)

x^{2} +18x + 81

factors to:

(x + 9) (x + 9)

because:

9 x 9 = 81

9 + 9 = 18

So:

It factors completely too:

6(x + 9) (x + 9)

or

6 (x+9)^{2}

6 0
3 years ago
Marco has a jug with 37 1/2 ounces of juice. He wants to use some small cups to pass out samples of the juice. Each cup can hold
Neko [114]

37 1/2 divided by 2 1/2=15

Marco can fill 15 cups with juice samples.

Check:
2 1/2x15=37 1/2
8 0
3 years ago
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