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coldgirl [10]
2 years ago
5

A projectile is fired directly upward. The compound inequality 89 – 9.8t < 10.6 or 89 – 9.8t > 20.4 represents a projectil

e’s velocity that is less than 10.6 m/s or greater than 20.4 m/s. Solve the compound inequality. The solution to the inequality 89 – 9.8t < 10.6 is ? . The solution to the inequality 89 – 9.8t > 20.4 is? . To determine when the projectile hits the ground, solve 89 – 9.8t = 0 for t. Rounding to the nearest whole second, t is about ? seconds. The viable solution set is ?
Mathematics
2 answers:
Tom [10]2 years ago
7 0

Given compound inequality 89 – 9.8t < 10.6 or 89 – 9.8t > 20.4.

Let us solve them one by one.

89 – 9.8t < 10.6

Subtracting 89 from both sides, we get

89-89 – 9.8t < 10.6 - 89

-9.8t < -78.4

Dividing both sides by -9.8, we get

t > 8.

<em>Note: Inequality sign get flip on dividing both side by a negative number.</em>

89 – 9.8t > 20.4

Subtracting 89 from both sides, we get

89-89 – 9.8t < 20.4 - 89

-9.8 t < - 68.6.

Dividing both sides by -9.8, we get

t < 7

<em>Note: Inequality sign get flip on dividing both side by a negative number.</em>

<h3><em> The solution to the inequality 89 – 9.8t < 10.6 is </em>t > 8. </h3><h3> The solution to the inequality 89 – 9.8t > 20.4 is t < 7.</h3>

When the projectile hits the ground:

89 – 9.8t = 0

Subtracting 89 from both sides, we get

89 - 89 – 9.8t = 0 -89.

-9.8t = -89.

Dividing both sides by 9.8, we get

<h3>to the nearest whole second  t is about 9 seconds.</h3><h3>Therefore, variable solution set is {t < 7, t > 8, t = 9}.</h3><h3 />
Nady [450]2 years ago
3 0

Answer:

The solution to the inequality 89 – 9.8t < 10.6 is ?

t > 8

The solution to the inequality 89 – 9.8t > 20.4 is?

t < 7

To determine when the projectile hits the ground, solve 89 – 9.8t = 0 for t.

t = 9.08163265306

Rounding to the nearest whole second, t is about ?

9 seconds.

The viable solution set is ?

[0, 7) or (8, 9]

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Option B: TU $\cong$ TS PU $\cong$ TU

Option C: The length of line segment PR is 13 units.

Explanation:

Given that the circle is inscribed in triangle PRT. Points Q, S, and U of the circle are on the sides of the triangle. Point Q is on side P R, point S is on side R T, and point U is on side P T.

The length of RS is 5, the length of PU is 8 and the length of UT is 6.

Option A: The perimeter of the triangle is 19 units.

The perimeter of the triangle is given by

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Since, P, T and R are tangents to the circle and we know that "Tangents to a circle drawn to a point outside the circle are equal in length".

Thus, we have,

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Substituting the values in the perimeter of ΔPRT, we get,

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Thus, the perimeter of the triangle is 38 units.

Hence, Option A is not the correct answer.

Option B : TU $\cong$ TS PU $\cong$ TU

Since, P, T and R are tangents to the circle and we know that "Tangents to a circle drawn to a point outside the circle are equal in length".

Then TU $\cong$ TS PU $\cong$ TU

Hence, Option B is the correct answer.

Option C: The length of line segment PR is 13 units.

The length of PR is given by

PR = PQ + QR

Substituting the values RQ = 5 and PQ = 8, we get,

PR = 5 + 8 = 13 units

Thus, the length of line segment PR is 13 units.

Hence, Option C is the correct answer.

Option D: The length of line segment TR is 10 units.

The length of TR is given by

TR = TS + SR

Substituting the values TS = 6 and SR = 5, we get,

TR = 6 + 5 = 11 units

Thus, the length of line segment TR is 11 units

Hence, Option D is not the correct answer.

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