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Ronch [10]
3 years ago
14

approximate each irrational number to the nearest hundredth without using a calculator square root of 118 and 319​

Mathematics
1 answer:
Strike441 [17]3 years ago
8 0

Answer:

\sqrt{118}\approx 10.86

\sqrt{319}\approx 17.86

Step-by-step explanation:

Consider the provided number.

We need to find the approximate value of \sqrt{118} to the nearest hundredth.

First find two perfect squares that the irrational number falls between.

100

118 is lying between 100 and 121, therefore the square root value of 118 will be somewhere between 10 and 11.

\sqrt{100}

10

118 is closer to 121 as compare to 100.

Therefore, \sqrt{118}\approx 10.86

Consider the number \sqrt{319}

First find two perfect squares that the irrational number falls between.

289

319 is lying between 289 and 324, therefore the square root value of 319 will be somewhere between 17 and 18.

\sqrt{289}

17

319 is closer to 324 as compare to 289.

Therefore, \sqrt{319}\approx 17.86

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What number must you add to the polynomial below to complete the square? x^2+17x
Juliette [100K]

The number added to the polynomial by completing the square is \frac{289}{4}

Explanation:

Given that the polynomial is x^{2} +17x

We need to determine the number that is added to the polynomial to complete the square.

The last term of the polynomial can be determined by dividing the term 17 by 2 and then squaring the term.

Thus, we have,

Last term = \frac{17}{2}

Now, squaring the term, we have,

Last term = (\frac{17}{2})^2=\frac{289}{4}

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Object is thrown upward from a height of 15 ft at an initial vertical velocity of 30 ft per second. How long will it take to hit
dem82 [27]

Answer:

2.25 s.

Step-by-step explanation:

We'll begin by calculating the time taken for the object to get to the maximum height from the point of projection. This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₁) to reach the maximum height from the point of projection =?

v = u – gt₁ (since the object is going against gravity)

0 = 30 – (32.15 × t₁

0 = 30 – 32.15t₁

Collect like terms

0 – 30 = – 32.15t₁

– 30 = – 32.15t₁

Divide both side by – 32.15

t₁ = –30 / –32.15

t₁ = 0.93 s

Next, we shall determine the maximum height reached by the object from the point of projection.

This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Maximum height (h) reached from the point of projection =?

v² = u² – 2gh (since the object is going against gravity)

0² = 30² – (2 × 32.15 × h)

0 = 900 – 64.3h

Collect like terms

0 – 900 = – 64.3h

– 900 = – 64.3h

Divide both side by – 64.3

h = –900 / –64.3

h = 14 ft

Thus, the maximum reached by the object from the point of projection is 14 ft.

Next, we shall determine the height to which the of object is located from the maximum height reached to the ground. This can be obtained as follow:

Height (h₀) from which the object was projected = 14 ft

Maximum Height (h) reached from the point of projection = 14 ft

Height (hₗ) to which the of object is located from the maximum to the ground =?

hₗ = h₀ + h

hₗ = 14 + 14

hₗ = 28 ft

Thus, the height to which the of object is located from the maximum reached to the ground is 28 ft.

Next, we shall determine the time taken for the object to get to the ground from the maximum height reached. This can be obtained as follow:

Height (hₗ) to which the of object is located from the maximum to the ground = 28 ft

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₂) taken for the object to get to the ground from the maximum height reached =?

hₗ = ½gt₂²

28 = ½ × 32.15 × t₂²

28 = 16.075 × t₂²

Divide both side by 16.075

t₂² = 28 / 16.075

Take the square root of both side

t₂ = √(28 / 16.075)

t₂ = 1.32 s

Finally, we shall determine the time take for the object to get to the ground from the point of projection. This can be obtained as follow:

Time (t₁) to reach the maximum height from the point of projection = 0.93 s

Time (t₂) taken for the object to get to the ground from the maximum height reached = 1.32 s

Time (T) take for the object to get to the ground from the point of projection =?

T = t₁ + t₂

T = 0.93 + 1.32

T = 2.25 s.

Therefore, the time take for the object to get to the ground from the point of projection is 2.25 s.

7 0
2 years ago
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