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cricket20 [7]
3 years ago
13

I need help ASAP!!!! Pleaseeee

Mathematics
1 answer:
poizon [28]3 years ago
5 0

Answer:

2nd and 3rd ordered pair are on the graph

Step-by-step explanation:

Substitute the x- coordinate into the equation and if the value obtained is equal to the y- coordinate then it lies on the graph.

(1, 625 )

f(1) = - 25^{1+1} = - 25^{2} = - 625 ≠ 625 ← No

(0, - 25 )

f(0) = - 25^{0+1} = - 25^{1} = - 25 ← Yes

(- 1, - 1 )

f(- 1) = - 25^{-1+1} = - 25^{0} = - 1 ← Yes [ using a^{0} = 1 ]

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xenn [34]

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6z + 7  = 8z - 9

8z - 9 = 6z + 7

Add sides 9

8z - 9 + 9 = 6z + 7 + 9

8z = 6z + 16

Subtract sides 6z

- 6z + 8z =  - 6z + 6z + 16

2z = 16

Divide sides by 2

\frac{2z}{2}  =  \frac{16}{2}  \\

z = 8

Thus ;

NP = 8z - 9

NP = 8(8) - 9

NP = 64 - 9

NP = 64 - 4 - 5

NP = 60 - 5

NP = 55

Done...

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6 0
3 years ago
A bathtub that holds 32 gallons of water contains 17 gallons of water. You begin filling it, and after 5 minutes, the tub is ful
Alik [6]
What’s the question???
8 0
3 years ago
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I need to know how to do this step by step
mariarad [96]
Multiply negative 0.5 times negative 8
The answer is 4
5 0
3 years ago
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What are the values of x and y in PQRS? What are PR and SQ?
mamaluj [8]

Step-by-step explanation:

sol;

x+1=y...(1)

3y-7=2x....(2)

or, 3(x+1)-72x [from (1)]

or, 3x+3-7=2x

or, 3x-2x=7-3

     x=4

now,

putting the value of x in (1)

y=x+1

 =4+1

 =5

PR and SQ are the diagonal of PQRS.

4 0
3 years ago
In one jar, I have two balls labelled 1 and 2 respectively. In a second jar, I have three balls labelled 0, 1 and 2 respectively
PolarNik [594]

Step-by-step explanation:

this is a kind of trick question, actually.

with whatever we draw, we produce X values as power of 3.

to be precise, we can have only

3⁰ = 1

3¹ = 3

3² = 9

3⁴ = 81

due to the possible combinations of drawn numbers (e.g. 3 cannot be created by a multiplication of 0s, 1s and 2s).

so, mostly, these results cannot be exact factors of 1024.

1024 cannot be divided by 3, nor by 9 nor by 81.

but 1024 is a multiple of 1 (as is every number).

so, we are looking at the probability to get 0 as multiplication result of the numbers on the 2 drawn balls.

the only possibilities are

1 and 0

2 and 0

out of in total 6 (2×3) different outcomes

1 and 0

1 and 1

1 and 2

2 and 0

2 and 1

2 and 2

the probability of this "0" event is again

number of desired outcomes / number of possible outcomes = 2/6 = 1/3

6 0
3 years ago
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