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LekaFEV [45]
2 years ago
15

Find the area of a rectangle with a perimeter of 12 meters and a base of 4 meters

Mathematics
1 answer:
stepladder [879]2 years ago
7 0
Considering the perimeter (P)
p = 2 \times b + 2 \times h
where b is the base and h is the height, then
12 = 2 \times 4 \times 2 \times h \\  \\ h =  \frac{12 - (2 \times 4)}{2}  \\  \\ h = 2
Being the area (A)
a = b \times h \\  \\ a = 4 \times 2 \\  \\ a = 8 {m}^{2}

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One more question on this test guys. please answer as fast as possible. If you do,
GaryK [48]

1. The given rectangular equation is x=2.

We substitute x=r\cos \theta.

r\cos \theta=2

Divide through by \cos \theta

r=\frac{2}{\cos \theta}

r=2}\sec \theta

\boxed{x=2\to r=2\sec \theta}

2. The given rectangular equation is:

x^2+y^2=36

This is the same as:

x^2+y^2=6^2

We use the relation r^2=x^2+y^2

This implies that:

r^2=6^2

\therefore r=6

\boxed{x^2+y^2=36\to r=6}

3. The given rectangular equation is:

x^2+y^2=2y

This is the same as:

We use the relation r^2=x^2+y^2 and y=r\sin \theta

This implies that:

r^2=2r\sin \theta

Divide through by r

r=2\sin \theta

\boxed{x^2+y^2=2y\to r=2\sin \theta}

4. We have x=\sqrt{3}y

We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta\sqrt{3}

This implies that;

\tan \theta=\frac{\sqrt{3}}{3}

\theta=\frac{\pi}{6}

\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}

5. We have x=y

We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta

This implies that;

\tan \theta=1

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

\boxed{x=y\to \theta=\frac{\pi}{4}}

6 0
3 years ago
Paige wants to sew lace around a round table cloth. The table cloth has a diameter of 6ft. About how much lace will Paige need?
Free_Kalibri [48]
It is B because u need enough for it to do it right
5 0
3 years ago
What is the max or min and range of y=2x^2+20x-12
shtirl [24]

Answer: its x = 3

Step-by-step explanation:

The maximum or minimum of a quadratic function occurs at

x

=

−

b

2

a

. If

a

is negative, the maximum value of the function is

f

(

−

b

2

a

)

. If

a

is positive, the minimum value of the function is

f

(

−

b

2

a

)

.

f

min

x

=

a

x

2

+

b

x

+

c

occurs at

x

=

−

b

2

a

Find the value of

x

equal to

−

b

2

a

.

x

=

−

b

2

a

Substitute in the values of

a

and

b

.

x

=

−

−

12

2

(

2

)

Remove parentheses.

x

=

−

−

12

2

(

2

)

Simplify

−

−

12

2

(

2

)

.

Tap for more steps...

x

=

3

The maximum or minimum of a quadratic function occurs at

x

=

−

b

2

a

. If

a

is negative, the maximum value of the function is

f

(

−

b

2

a

)

. If

a

is positive, the minimum value of the function is

f

(

−

b

2

a

)

.

f

min

x

=

a

x

2

+

b

x

+

c

occurs at

x

=

−

b

2

a

Find the value of

x

equal to

−

b

2

a

.

x

=

−

b

2

a

Substitute in the values of

a

and

b

.

x

=

−

−

12

2

(

2

)

Remove parentheses.

x

=

−

−

12

2

(

2

)

Simplify

−

−

12

2

(

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x

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4 0
2 years ago
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Usimov [2.4K]
The answer is D. Group =2
8 0
3 years ago
An opaque bag contains 5 green marbles 3 blue marbles and 2 red marbles. What is the probability of randomly picking out a marbl
anastassius [24]

Answer= .2 probability

Explanation: Since there is a total of 10 marbles, and 2 are red, you would divide the red marbles (2) out of the total marbles (10) which would get u .2. Since it’s asking for probability, we would keep .2 a decimal.

6 0
2 years ago
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