Answer:
Possible locker numbers for Selma are: 3, 5 and 29
Step-by-step explanation:
First of all, let us have a look at the definition of a prime number.
A <em>prime number</em> is a number which is divisible either by 1 or the number itself.
No other number divides the prime number other than 1 and the number itself.
Now, let us factorize the given number 435 and let us observe the what all possibilities are there.
Factorizing 145 further:
No further factors are possible.
Therefore 3, 5 and 29 are the prime factors of 435.
So, the answer is:
Possible locker numbers for Selma are: 3, 5 and 29
20x+50=450
-50 -50
20x=400
Divide both sides by 20
X=20
Both expression have the same denominator: 9x²-1. Thus it must not be 0.
9x²-1=(3x-1)(3x+1)=0, resulting x=+-1/3.
Restrictions: x in R\{-1/3, 1/3}
Adding those expressions:
E=(-x-2)/(9x²-1 ) + (-5x+4)/(9x²-1)=
(-x-2-5x+4)/(9x²-1)=(-6x+2)/(9x²-1)=
(-2)(3x-1)/(9x²-1)=-2/(3x+1)
E=-2/(3x+1)
Answer:
1. 5/3 cm
2. 10 in
Step-by-step explanation:
1. The length of a rectangle is 4 cm more than 3 times its width.
Let the length be L and the width B
Then,
L = 4 + 3B
If the area of the rectangle is 15 cm2
Area = LB = 15
B(4 +3B) = 15
3B² + 4B - 15 = 0
3B² - 5B + 9B - 15 = 0
B(3B - 5) + 3(3B - 5) = 0
(3B - 5)(B + 3) = 0
3B - 5 = or B + 3 = 0
B = 5/3 or -3
since B cannot be negative, B = 5/3 cm
2. The ratio of length to width in a rectangle is 2 ∶ 3.
Let the length be L and the width B
Then,
L/B = 2/3
2B = 3L, B = (3/2)L
when the area is 150 in2.
Area = LB = 150
(3/2)L² = 150
L² = 150×2÷3
L² = 100
L = √100
L = ±10
since the length cannot be negative, L = 10 in
