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3 years ago

The area of a circle is 113.04 square meters. What is the circle's radius? (Using 3.14 as pi)

Mathematics

2 answers:

notka56 [123]3 years ago

4 0

Answer:

Step-by-step explanation:

113,04=3,14×r²

113,04/3,14=r²

*√ both sides*

julia-pushkina [17]3 years ago

3 0

Answer:

R=6 PLS GIVE ME BRAINLIST

Step-by-step explanation:

113.04=3.14R^2

36=R^2

R=6

WE HAVE to work backwards so we first plug in the area and then divide 3.14 by the area so we get 36 and then find a number tht squares to 36 which is 6

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Find the end behavior of f(x)= x^2+6x+6

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falls to the left and rises to the right

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3 years ago

Can someone help me with this diagram problem?

mart [117]

Answer: the diagonal is 10 units

Step-by-step explanation:

The rectangle has 2 parallel and equal sides. Each of the 4 angles in a rectangle is 90°. Therefore, the diagonal of the rectangle divides it into 2 equal right angle triangles and the diagonal represents the hypotenuse of both right triangles.

To determine the length of the diagonal of the rectangle, d, , we would apply Pythagoras theorem which is expressed as

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6 0

3 years ago

A projectile is fired from ground level with an initial speed of 550 m/sec and an angle of elevation of 30 degrees. Use that the

Nana76 [90]

Answer:

x (max) = 26760 m

y (max) = 3859 meters

V = 549.5 m/sec

Step-by-step explanation:

Equations to describe the projectile shot movement are:

a(x) = 0 V(x) = V(₀) *cos α x = V(₀) *cos α * t

a(y) = -g V(y) = V(₀) * sin α - g*t y = V(₀) * sin α *t - (1/2)*g*t²

a ) What is the range of the projectile. α = 30°

then sin 30° = 1/2 cos 30° = √3 /2 and tan 30° = 1/√3

x maximum occurs when in the equation of trajectory we make y = 0

Then

y = x*tan α - g*x / 2*V(₀)²*cos² α

x*tan α = g*x / 2* V(₀)²*cos² α

By subtitution

1/√3 = 9.8* x(max) / 2* (550)²*0.75

(1/√3) * 453750 / 9.8 = x (max)

x (max) = 453750 / 16.95 meters

x (max) = 26760 m

The maximum height is when V(y) = 0

We compute t in that condition

V(y) = 0 = V(₀) * sin α - g*t

t = V(₀) * sin α / g ⇒ t = 550* (1/2) / 9.8

t = 28.06 sec

Then h (max) = y(max) = V(₀) sin α * t - 1/2 g* t²

y (max) = 550* (1/2)*28.06 - (1/2)* 9.8 * (28.06)²

y (max) = 7717 - 3858

y (max) = 3859 meters

What is the speed when the projectile hits the ground

V = V(x) + V (y) and t = 2* 28.06 t = 56.12 sec

mod V =√ V(x)² + V(y)²

V(x) = V(₀) cos α = 550 * √3/2

V(x) = 475.5 m/sec V(x)² = 226338 m²/sec²

V(y) = 550*1/2 - 9.8* 56.12 ⇒ V(y) = 275 - 549.98

V(y) = - 274.98 V(y) ² =

V = √ 226338 + 75614 ⇒ V = 549.5 m/sec

7 0

3 years ago

The function f(t) = -16t2 + 64t + 5 models the height of a ball that was hit into the

koban [17]

The best statement that compares the length of time each ball is in the air is seen in the second option.

<h3>What is the function of a quadratic equation?</h3>

A quadratic function follows the condition: f(x) = ax² + bx + c. Here (a, b, and c) are real numbers and a ≠ 0.

From the given information:

The function f(t) = -16t² +64t + 5 models the height of a ball

Where;

time (t) ⇒ seconds
height (h) ⇒ feet

The table given is as follows:

Time (in seconds) height, g(t) in feet

0 4

1 36

2 36

3 4

From the table, when t = 0, the function becomes:

f(t) = -16t² +64t + 5

f(t) = -16(0)² + 64(0) + 5

f(t) = 5

Therefore, we can conclude that the best statement that compares the ball is: f(t) is in the air for about 5 seconds, and the ball represented by g(t) is in the air for about 3 seconds.

Learn more about the function of a quadratic equation here:
brainly.com/question/14712068

5 0

2 years ago