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Drupady [299]
3 years ago
13

Two fair dice are rolled. Determine the following probability

Mathematics
1 answer:
VashaNatasha [74]3 years ago
8 0
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ZC and ZD are vertical angles with mZC = -3x+58 and mZD= x - 2.
kogti [31]

Answer:

×=-2

0=×-2

×=-2

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3 0
3 years ago
Express this as a difference: 7+3<br> Plz explain If you do you will get brainlyes
likoan [24]
10 because if you 3 with 7 I will make 10




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Simplify 6x•1/x^-5•x^-2
sukhopar [10]

Answer:

6x^4

Step-by-step explanation:

5 0
3 years ago
Need help solving algebraically!
alekssr [168]
We need to account for both x values on either side of the length, and width.
Thus, the length becomes 10 + x + x = 10 + 2x
and the width becomes 5 + x + x = 5 + 2x

For the second question, I'm assuming we don't account for the area that is covered by the garden.
Then we can say that the path is measured by: (5 + 2x)(10 + 2x) - 50, which is the area of the garden itself.

(5 + 2x)(10 + 2x) - 50 = 54

Expanding the brackets:
50 + 10x + 20x + 4x^{2} - 50 = 54
4x^{2} + 30x = 54
2x^{2} + 15x - 27 = 0
2x^{2} - 3x + 18x - 27 = 0
x(2x - 3) + 9(2x - 3) = 0

(x + 9)(2x - 3) = 0
x = -9, or x = 3/2

Since x > 0, then x ≠ -9
Thus, the only x-value we can take is x = 3/2
6 0
3 years ago
Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one sam
kogti [31]

Answer:

Advance tickets-$15

Same-day tickets-$20

Step-by-step explanation:

let x be the cost of advance tickets and y cost of same-day tickets:

x+y=35\ \ \ \ \  \ \ ...i

Given that there were 40 advance and 25 same-day tickets for a total of $1100:

40x+25y=1100\\\\8x+5y=220\ \ \ \  \ \ \ \ \ \ \ ...ii

#Make x the subject in i and substitute in ii:

x=35-y\\\\\therefore 8x+5y=220, x=35-y\\\\8(35-y)+5y=220\\\\280-8y+5y=220\\\\60=3y\\\\y=20\\\\x=35-y=35-20=15

Hence, advance tickets cost $15 each while same-day tickets cost $20 each.

5 0
3 years ago
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