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gladu [14]
3 years ago
11

Which ordered pairs show the location of the bleachers? check all that apply.

Mathematics
1 answer:
just olya [345]3 years ago
7 0
It’s (b) (1.50,2.0)
hope this helped :))
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Find the area of the circle. Use 3.14 for Pi
IceJOKER [234]

Answer:

113.04 cm squared

Step-by-step explanation:

the area of a circle is:

A=πr2

so as you written before the pi must be 3.14

and the radius is the diameter/2(12/2=6)

A= 3.14 x 6 x 6

A= 113.04 cm squared

3 0
3 years ago
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Find the area of a rhombus with side length 6 and an interior angle with measure $120^\circ$.
Serhud [2]

Answer:

The area of a rhombus is 18\sqrt{3} square units.

Step-by-step explanation:

Side length of rhombus = 6 units.

Interior angle of rhombus = 120°

another Interior angle of rhombus = 180°-120° = 60°.

Draw an altitude.

In a right angled triangle

\sin \theta=\frac{opposite}{hypotenuse}

\sin (60)=\frac{h}{6}

\frac{\sqrt{3}}{2}=\frac{h}{6}

Multiply both sides by 6.

3\sqrt{3}=h

The height of the rhombus is 3\sqrt{3}.

Area of a rhombus is

Area=base\times height

Area=6\times 3\sqrt{3}

Area=18\sqrt{3}

Therefore, the area of a rhombus is 18\sqrt{3} square units.

8 0
3 years ago
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Find the interest using the formula I = Prt Principal: $1800 Interest Rate: 6.5% Time: 30 months Interest =​
Oliga [24]
I = P r t

I = (1800) (6.5÷100) (30÷12)

I = (1800) ( 0.065) (2.5)

I = 292.50$
8 0
3 years ago
A piece of driftwood floats on the surface of the ocean. Due to the wave motion, its height rises and falls relative to the heig
Black_prince [1.1K]
Assuming that both the max and min heights of the wave are multiplied by 4, the new max height is 2*4 =8ft, and the min height is -1.5*4 =-6ft.

<span>What is the minimum height of the driftwood in the storm? -6ft.
What is the distance between the maximum and minimum heights of the driftwood during the storm? 8ft-(-6ft)=14 ft. </span>

3 0
3 years ago
In your BIO 111 course, you learn that human hair can vary in diameter, but one estimate of the average diameter is 50 microns o
Dafna11 [192]

Answer:

5000 particles of size 10 nanometer should be stacked

Step-by-step explanation:

Average diameter of the of the human hair = 50 micron = 50 × 10⁻⁶ m

Now,

let the number of 10 nanometer i.e 10 × 10⁻⁹ m particle be 'n'

thus,

n × 10 × 10⁻⁹ m = 50 × 10⁻⁶ m

Therefore to find the number of 10 nanometer i.e 10 × 10⁻⁹ m particle,

n = \frac{50\times10^{-6}}{10\times10^{-9}}

or

n = 5000

Hence,

5000 particles of size 10 nanometer should be stacked

4 0
3 years ago
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