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dangina [55]
3 years ago
11

Which procedure justifies whether -3x(5-4)+3(x-6) is equivalent to -12x-6?

Mathematics
2 answers:
NARA [144]3 years ago
8 0

Option (a) is correct.

Step-by-step explanation:

Given : Expressions 3x(5-4)+3(x-6) and -12x-6

We have to choose which option from the given options justifies the correct procedure to show the given two expressions are equivalent or not.

Since to check two given expressions are equivalent or not. Simply put the same value of unknown and evaluate the values of both the given expressions. If it comes out to be same then they are equivalent otherwise not.

Consider the given expression 3x(5-4)+3(x-6) and -12x-6

Evaluate both at x = 2

So, 3(2)(5-4)+3(2-6) = -18

and  -12(2)-6 = -30

So both the expressions are not equivalent.

Hence, option (a) is correct.

DENIUS [597]3 years ago
4 0

Answer:

Step-by-step explanation:

-3x(5-4) + 3(x-6)

-15x + 12 + 3x - 18

-15x + 3x +12 - 18

-12x - 6

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Gretchen made a paper cone to hold a gift for a friend. The paper cone was 14 inches high and had a
LUCKY_DIMON [66]

Answer:

268.0 in²

Step-by-step explanation:

refer to attached graphic as reference

volume of cone, V = (1/3) πr²h

in our case, we are given r = 4" and h = 16"

substituting this into equation:

V = (1/3) πr²h

=  (1/3) ·(3.14) · (4)²· (16)

= 267.94667 in²

= 268.0 in² (nearest tenth)

8 0
2 years ago
Read 2 more answers
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
Solve for x d(-3+x)= kx+9
olga nikolaevna [1]
1. Write the equation:
d(-3 + x) = kx + 9
2. "Open" the parenthesizes:
-3d + dx = kx + 9
3. Take kx to the left side, -3d to the right, everything with different sign (+ replace with - when transferring, - replace with +):
dx - kx = 9 + 3d
4. Factor x in the left side:
x(d - k) = 9 + 3d
5. Finally, divide the whole equation by (d - k):
x = (9 + 3d)/(d - k)
That's the final answer. Good luck!
4 0
3 years ago
Read 2 more answers
I need help with 8 9 thanks
sasho [114]
#8: $900. (I found this answer by multiplying 3000 and 0.3 by the way)
#9: 160 miles. I found this answer by multiplying 80 by two, since two halves/50% make a whole or 100%.
3 0
3 years ago
mia wants to earn 2500 in her job at a bycicle store she is paid 55 for every bycicle she tunes up the minimum number of bycicle
Thepotemich [5.8K]
Answer: 46

Explanation: If Mia is paid $55 for every bicycle, she would need to fix up/sell 46 bicycles. This is because 45 x 55 = 2475 and Mia wants over 2500 but 46 x 55 = 2530. Therefore, 46 is the minimum number of bicycles.
5 0
3 years ago
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