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Mrrafil [7]
3 years ago
15

A parallelogram is being painted on the wall of a playroom. The parallelogram measures 7.3 meters in length and has a height of

5 meters. How many square meters of paint are needed?
Mathematics
1 answer:
iris [78.8K]3 years ago
6 0

Answer: 36.5 m²

Step-by-step explanation:

Whenever a unit is square, it means that it is measuring Area. Square meters are therefore used to denote the area of a shape.

This question is therefore asking for the area of the Parallelogram which is calculated by the formula:

= Length * Width

= 7.3 * 5

= 36.5 m²

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Komok [63]

Answer:

He has to work for more than 8 hours.

Step-by-step explanation:

8t + 4 > 52

8t + 4 - 4 > 52 - 4

8t > 48

8t/8 > 48/8

t > 8

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3 years ago
HELP PLEASE
olga55 [171]
The slope of a line includes a x and y axis, or y/x. Your two answers are -7/4 and -8/5. Hope that helps!
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4x+12=-7y<br> -y+12=4x<br><br> PLS HELP!!!!<br> Addition/ Elimination method
hoa [83]
4×+7y=-12 __ -4x-y=-12 ________ 0x+6y=-24 , so you eleminated by it self 6y/6=-24/6 , you get : y=-4 and then you plug in 4x+7(-4)=-12 , that's your answer 4x-28 =-12 , you divided in both side 4x=-12+28 , 4x=16 , and then x=16/4 =4 , so x=4
3 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
Which are steps in the process of completing the square used to solve the equation 3 – 4x = 5x2 – 14x? Check all that apply.
avanturin [10]
Add 4x over to 14x, 3=5x2- 18x then add 18x to the other side 3+ 18x= 5x2then divided each side by 5,( 3+18x/5=x2) then square root each side the answer to x would be the square root of 3+18x/5
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3 years ago
Read 2 more answers
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