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cestrela7 [59]
3 years ago
5

The sum of 4 squared and a number t

Mathematics
2 answers:
drek231 [11]3 years ago
7 0

Answer:

16+t

4^2+t would give you this answer

BartSMP [9]3 years ago
3 0

Answer:

4^2 + t

Step-by-step explanation:

<u>Step 1:  Change words into an expression</u>

Sum of 4 squared and number t

4^2 + t

Answer:  4^2 + t

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At Westside middle school there are 42 classrooms with a 28 desks in each. About how many desks are there?
Marianna [84]
42*28=1176.
There are 1176 desks
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3 years ago
If Roger were 32 years older, he would be three times as old as he is now. How old is<br> Roger?
stellarik [79]

Answer:

16

Step-by-step explanation:

Set up the equation ->

x+32=3x  

(you can get this easily by writing the equation down as you read the problem)

--> revisiting  x+32=3x

-x on both sides to isolate the number and put the like terms together

We are left with  32=2x

Divide 2 on both sides, and you will get x (his current age) --> x=16

Roger is currently 16 years old.

6 0
2 years ago
Read 2 more answers
What is the approximate volume of a cylinder with a height of 2 ft and a radius of 6 ft?
patriot [66]
We know, Volume of a Cylinder = πr²h
Here, r= 6 & h = 2

Substitute their values into the expression:
V = 3.14 * (6)² * 2
V = 3.14 * 36 * 2
V = 3.14 * 72
V = 226.08

In short, Your Answer would be 226.08 Ft³

Hope this helps!
5 0
3 years ago
On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or
almond37 [142]

Answer:

a) \vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j, b) \theta = \frac{\pi}{4}, c) y_{max} = 84.375\,m, t = 3.75\,s.

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j

The particular expression for the cannonball is:

\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j

b) The components of the position of the cannonball before hitting the ground is:

x = (90\cdot \cos \theta)\cdot t

0 = 90\cdot \sin \theta - 6\cdot t

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta}  \right)

0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x

0 = 4050\cdot \sin 2\theta - 6\cdot x

6\cdot x = 4050\cdot \sin 2\theta

x = 675\cdot \sin 2\theta

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

\frac{dx}{d\theta} = 1350\cdot \cos 2\theta

1350\cdot \cos 2\theta = 0

\cos 2\theta = 0

2\theta = \frac{\pi}{2}

\theta = \frac{\pi}{4}

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta

\frac{d^{2}x}{d\theta^{2}} = -2700

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

y = 45\cdot t - 6\cdot t^{2}

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

\frac{dy}{dt} = 45 - 12\cdot t

45-12\cdot t = 0

t = \frac{45}{12}

t = 3.75\,s

Now, the second derivative is used to check if such solution leads to a maximum:

\frac{d^{2}y}{dt^{2}} = -12

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}

y_{max} = 84.375\,m

7 0
3 years ago
Find the 11th term of the geometric sequence 6, -18, 54
Nataly [62]
I don’t feel like doing it but I we’ll tell you how to do it you multiply it by 3 so 6 x 3 equals 18 just in negative do 18 x 3 equal 54 so every other on is a negative it’s Simone
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