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Solnce55 [7]
3 years ago
15

Is it possible to find the prime factorization of 29 explain why or why not

Mathematics
2 answers:
zhuklara [117]3 years ago
6 0

Answer:

  • No

Step-by-step explanation:

  • 29 is a prime number and therefore you can't factorize it.
zloy xaker [14]3 years ago
5 0

Answer:

No

Step-by-step explanation:

  • 29 is a prime
  • It has only two factors.
  • i.e 1 and the number itself 29.
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Is 4.9199199919999... a rational or irrational number?
ratelena [41]

Answer:

irrational

Step-by-step explanation:

rational is if its repeating like (1.11111111) irrational if its (1.2464383252222)

7 0
3 years ago
Solve for x ....... thanks !
Len [333]
If we can match teh bases we can solve
because if x=x and xᵃ=xᵇ, we can conclude that a=b

16=2⁴
32=2⁵
rememeber that (x^m)^n=x^{mn}

16^{3x+2}=32^{-2x-7}
(2^4)^{3x+2}=(2^5)^{-2x-7}
2^{4(3x+2)}=2^{5(-2x-7)}
2=2 so we conclude that 4(3x+2)=5(-2x-7)

4(3x+2)=5(-2x-7)
expand/distribute
12x+8=-10x-35
add 10x both sides
22x+8=-35
minus 8 both sides
22x=-43
divide both sides by 22
x=-43/22
8 0
3 years ago
Is -77 a rational or irrational number?​
Butoxors [25]

It’s a rational number.

Hope this helps!

7 0
4 years ago
Read 2 more answers
In a class of 29 students, 19 are female and 14 have an A in the class. There are 8students who are male and do not have an A in
masya89 [10]

SOLUTION

The total number of females =

19

if 14 have an A in the class, the number of students without A is:

29-14=15

8 male students do not have an A, therefore the number of female students without an A is:

15-8=7

The probability that a student does not have an A given that the student is female can be calculated thus:

\frac{\text{Total number of female students without an A}}{Total\text{ number of female students in the class}}\frac{7}{19}

8 0
1 year ago
Sebastian solved the radical equation y + 1 = but did not check his solution. (y + 1)2 = y2 + 2y + 1 = –2y – 3 y2 + 4y + 4 = 0 (
Softa [21]

Answer:

There are no true solutions to the equation.

Step-by-step explanation:

<u><em>The correct equation is</em></u>

y+1=\sqrt{-2y-3}

Solve for y

squared both sides

(y+1)^2=(-2y-3)

(y^2+2y+1)=(-2y-3)

y^2+2y+1+2y+3=0

y^2+4y+4=0

(y+2)(y+2)=0

y=-2

<em>Verify</em>

substitute the value of y in the original expression

-2+1=\sqrt{-2(-2)-3}

-1=1 ----> is not true

therefore

There are no true solutions to the equation.

4 0
3 years ago
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