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Korvikt [17]
3 years ago
6

Perimeter of 60 , length is 3 times the width, what are the dimensions

Mathematics
1 answer:
Murljashka [212]3 years ago
7 0
Length - l

Width - w

Given,

Perimeter = 60 

2(l +w) = 60

l + w = 30 

l = 30 - w

Given,

l = 3w

30 - w = 3w

4w = 30

--- w = 7.5

--- l = 3w = 3 * 7.5 = 29.5

Hence, the length of the rectangle is 29.5 and the width is 7.5
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Factor the trinomial by grouping.<br> 9x^3 - 3x^2 -30x
katen-ka-za [31]

Answer:

3x(3x+5)(x-2)

Step-by-step explanation:

9x^3-3x^2-30x= \\\\3x(3x^2-x-10)= \\\\3x(3x^2+5x-6x-10)= \\\\3x(x(3x+5)-2(3x+5))= \\\\3x(3x+5)(x-2)

Hope this helps!

4 0
3 years ago
A model rocket is launched with an initial velocity of 200 ft per second. The height h, in feet, of the rocket t seconds after t
Solnce55 [7]

Answer:

2.10 s, 10.40 s.

Step-by-step explanation:

We know that the height of the rocket is given by the function:

h=-16t^2+200t

We are asked to find the time for which the height of the rocket will be 350 ft. So, for that moment, we know the height but we don't know the time; however, we know that the equation can help us to find the time, doing h=350:

350=-16t^2+200t

The last is a quadratic equation, which can be put in the form at^2+bt+c=0 and solved applying the formula:

t=\frac{-b+-\sqrt{b^2-4ac} }{2a}

So, let's put the equation on the form at^2+bt+c=0 adding 16t^2 and subtracting 200t to each side of the equation; the result is:

16t^2-200t+350=0

So, we note that a=16, b=-200, and c=350.

Then,

t_1=\frac{200-\sqrt{200^2-4*16*350} }{2*16}=2.10

t_2=\frac{200+\sqrt{200^2-4*16*350} }{2*16}=10.40

According to the equation, that are the times for which the height will be 350 ft; that is because the rocket is going to ascend and then to fail again to the ground.

4 0
3 years ago
PLEASE TELL ME THE ANSWER TO PERCENTS BRAINPOP
coldgirl [10]
I wish I could tell you, I’m stuck on it too
5 0
2 years ago
Read 2 more answers
How many sides does a cube have
Lostsunrise [7]
A cube has 6 sides on it. Just think of a dice.

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5 0
3 years ago
Read 2 more answers
Someone help me please:)
Kryger [21]

Answer:

a. The time to strike the ground is 5 seconds

b. The time for the ball to be over than 96 ft is 1 second

Step-by-step explanation:

a. When the ball thrown vertically up ward to reach its maximum height at velocity = 0 then complete its motion vertically down ward to strike the ground its displacement = zero

∴ s(t) = 0 ⇒ 80t-16t^{2}=0

∴16t(5 - t) = 0

5 - t = 0⇒ ∴ t = 5 seconds

b. To tind the time that the ball will be over 96 ft over the ground substitute in s(t)=80t-16t^{2}

96=80t-16t^{2}

16t^{2}-80t+96=0

16(t^{2}-5t+6)=0

16(t-3)(t-2)=0

t = 3 sec. and t = 2 sec.

The ball will be at height 96 ft at time 2 sec. and 3 sec.

So the ball will be at height over than 96 = 3 - 2 = 1 sec.

8 0
2 years ago
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