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zvonat [6]
3 years ago
10

In a certain high school, 240 students made the honor roll

Mathematics
1 answer:
Sloan [31]3 years ago
8 0

could you finish the question? we cannot help you if we don't get the full question!

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Find the volume of the sphere.<br> Either enter an exact answer in terms of pi or use 3.14 for pi
Ratling [72]

Answer:

volume of sphere= 4/3πr³

<u>4</u>×<u>2</u><u>2</u>×10×10×10

3. 7

<u>88×1000</u>

21

<u>88000</u>

21

4190.48

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3 years ago
Which number is the largest? 7.2 ⋅ 10−6, 3.09 ⋅ 103, 2.04 ⋅ 104, 5 ⋅ 103
Tems11 [23]
The equation whit the largest outcome is 5*103=515
7 0
3 years ago
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2. The equation h(t)=−16t2+19t+110 gives the height of a rock, in feet, t seconds after it is thrown from a cliff.
Paul [167]
For this case we have an equation of the form:
 h (t) = - (1/2) * a * t ^ 2 + vo * t + h0
 Where,
 vo: initial speed
 a: acceleration:
 h0: initial height.
 We have the following equation:
 h (t) = - 16t2 + 19t + 110
 Therefore, the initial velocity is:
 vo = 19 feet / s
 Answer:
 
The initial velocity when the rock is thrown:
 
vo = 19 feet / s
8 0
3 years ago
Log(5) + log(3) rewrite the following in the form log(c).
Dennis_Churaev [7]
Log (5) + log (3) = log (15)
3 0
3 years ago
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If QS bisects PQT, SQT=(8x- 25),PQT=(9x+34), and SQR=112, find each measure.
Goryan [66]

The values of the angles are; x = 12°, m∠PQS = 71°, m∠PQT = 142° and m∠TQR = 41°

<h3>What are the measure of the each angle?</h3>

The required angles; x, m∠PQS, m∠PQT, and m∠TQR

The given parameters are bisects ∠PQT

m∠SQT = (8·x - 25)°

m∠PQT = (9·x + 34)°

m∠SQR = 112°

We have;

m∠PQT = m∠SQT + m∠PQS (Angle addition postulate)

m∠SQT ≅ m∠PQS (Angles formed by angle bisector are congruent)

m∠SQT = m∠PQS  by Definition of congruency

m∠PQT = 2 × m∠SQT

Therefore;

(9·x + 34)° = 2 × (8·x - 25)° = (16·x - 50)°

Collecting like terms gives:

(34 + 50)° = 16·x - 9·x = 7·x

7·x = 84°

x = 84°/7 = 12°

x = 12°

m∠SQT = (8·x - 25)°

Therefore;

m∠SQT = (8 × 12 - 25)° = 71°

m∠SQT = 71°

m∠PQT = 2 × m∠SQT

∴ m∠PQT = 2 × 71° = 142°

m∠PQT = 142°

m∠PQS = m∠SQT (Angles formed by the same bisector )

∴ m∠PQS = m∠SQT = 71°

m∠PQS = 71°

m∠SQR = m∠SQT + m∠TQR (Angle addition postulate)

m∠SQT = 71°

∴  m∠SQR = 112° = 71° + m∠TQR

m∠TQR = 112° - 71° = 41°

m∠TQR = 41°

Learn more about angles here:

brainly.com/question/2882938

#SPJ1

4 0
2 years ago
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