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lorasvet [3.4K]
3 years ago
8

U have to pick ALL the ones that are equivalent!

Mathematics
2 answers:
Ksju [112]3 years ago
7 0

Answer:

A. p -2/3p

B. 1/3p

E. 3p/9

Nutka1998 [239]3 years ago
5 0

A , B , E should be it

You might be interested in
PLS HELP ASAP ILL GIVE BRAINLKEST THANKS
natka813 [3]
The answer to this graph is (4,5)
4 0
2 years ago
Part A: Find the LCM of 7 and 12. Show your work. (3 points)
inna [77]

Answer:

A. 84; B. 8; C. 8 × 19

Step-by-step explanation:

Part A. Least common multiple

Step 1. List the prime factors of each.

7 = 7

12 = 2 × 2 × 3

Step 2: Multiply each factor the greatest number of times it occurs in either number.

7 has one 7; 12 has two 2s and one 3.

LCM = 7 × 2 × 2 × 3

LCM = 7 × 12

LCM = 84

Part B. Highest common factor

Find all the factors of 56 and 96.

Factors of 56: 1, 2, 4,     7, 8,      14,          28

Factors of 96: 1, 2, 4, 6,     8, 12,    16, 24,     32, 48

The highest factor that in both 56 and 96 is 8.

Part C. Factoring

56 + 96 = 8(7 + 12) = 8 × 19

The GCF is 8.

19 = 7 + 12 is the sum of two numbers that do not have a common factor.

4 0
4 years ago
What is the inverse of f(x) = 8x - 7
nignag [31]
Y=(x+7)/8

Work:
Y=8x-7 —> x=8y-7
Solve for y
x+7=8y
y=(x+7)/8

3 0
3 years ago
What is the value of x in -4x + 56 = 24
iren [92.7K]
-4x+56=24
-56=-56
———————
-4x =-32


-32/-4x
X=8
7 0
3 years ago
The product of two consecutive positive integers is 1111 more than their sum. find the integers.
Alex787 [66]
The correct question is
<span>The product of two consecutive positive integers is 11 more than their sum. find the integers.

Let
x-------> the first </span>positive integer
x+1-----> the second consecutive positive integer

we know that
x*(x+1)=11+(x+(x+1))----> x²+x=11+(2x+1)----> x²+x=2x+12
x²-x-12=0

using a graph tool------> to resolve the second order equation
see the attached figure

the solution is
x=4

therefore

the numbers are
x=4
x+1=5

the answer is
the numbers are 4 and 5

5 0
3 years ago
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