What is the greatest common factor of 15x^3y^2 and 20x^4y^4?
1 answer:
<h3>What is the greatest common factor of 15x³y² and 20x⁴y⁴? </h3>
<em>First we have to find the factors of 15x³y²</em>
<em>15x³y² = 3 * 5x³y² </em>
<em>Then we have to find the factors of 20x⁴y⁴</em>
<em>20x⁴y⁴ = 2²x¹y² * 5x³y²</em>
<em>Now we have to find the common factors to both numbers.</em>
<em>The common factors are </em><em>5x³y²</em>
Answer : GCF = (15x³y²,20x⁴y⁴) = 5x³y²
Hope this helps!
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Answer: 299
Step-by-step explanation:
Given
Family removes one-third of tress in June as much as they dd in July
They remove twice the number of July in august
Suppose they removed x trees in July
for June it is,
for August, it is
If they removed less than 1000 tress in 3 months
So, maximum number of tress removed in July is 299
Answer:
Therefore after 1 second of lunched the object reach a height of 16 feet.
Step-by-step explanation:
When a particle is thrown vertically upward direction on that time only gravitation force is worked on it. In that case the acceleration of the particle is - g.Since gravitation force works downward direction and the displacement of the particle is upward direction that's why acceleration is negative.
When particle comes down the the acceleration of the particle is g.
Since the displacement and acceleration work on the particle in same direction.
The value of g is 9.8 m/s².
Given that, the height of the particle can measure from the equation
Where t is time in second
To find out the time when the object reach a height 16 feet. We have put the value of h in the given equation.
Putting h=16 in the given equation
16= -16 t²+32
⇒16t²-32t+16=0
⇒16(t²-2t+1)=0
⇒t²-2t+1=0² [ ∵16≠0]
⇒(t-1)²=0
⇒t-1=0
⇒t=1
Therefore after 1 second of lunched the object reach a height of 16 feet.