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Fittoniya [83]
3 years ago
11

Brielle exercises for 3/4 hour each day for 6 days in a row. Altogether how many hours does she exercise during the 6 days?

Mathematics
2 answers:
mash [69]3 years ago
7 0
3/4 hour = 45 minutes
45 × 6 = 270 minutes
270 ÷ 60 = 4.5 hours
4.5 hours = 4 hours 30 minutes = 4 1/2 hours
Dmitry_Shevchenko [17]3 years ago
4 0

Answer:

4.5 hours does she exercise during the 6 days

Step-by-step explanation:

Unit rate defined as the rates are expressed as a quantity as 1 such as 2 meter per second or 4 kilometer per hour.

As per the statement:

Brielle exercises for 3/4 hour each day

⇒\text{Unit rate per day} = \frac{3}{4} = 0.75 hours

We have to find the how many hours does she exercise during the 6 days.

\text{Number of hours } = \text{Unit rate per day} \times 6

Substitute the given values we have;

\text{Number of hours } = 0.75 \times 6 = 4.5

Therefore, 4.5 hours does she exercise during the 6 days

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Answer:

$1343

Step-by-step explanation:

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x= 1580 (1-0.15)

x represents how much he actually pays.

1-0.15=0.85.

1580×0.85=1343

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2 years ago
Write the equation of a line that is perpendicular to x=-6x=−6x, equals, minus, 6 and that passes through the point (-1,-2)(−1,−
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Answer:

  y = -2

Step-by-step explanation:

The given line is a vertical line, so a perpendicular line will be a horizontal line. It will have an equation of the form ...

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3 years ago
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Question 14(Multiple Choice Worth 1 points)
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2 years ago
Calculus application
Ugo [173]

Answer:

49m/s

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Step-by-step explanation:

Given that :

Distance (s) = 178 m

Acceleration due to gravity (a) = g(downward) = 9.8m/s²

Velocity (V) after 5 seconds ;

The initial velocity (u) = 0

Using the relation :

v = u + at

Where ; t = Time = 5 seconds ; a = 9.8m/s²

v = 0 + 9.8(5)

v = 0 + 49

V = 49 m/s

Hence, velocity after 5 seconds = 49m/s

b) How fast is the ball traveling when it hits the ground?

V² = u² + 2as

Where s = height = 178m

V² = 0 + 2(9.8)(178)

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8 0
3 years ago
Will award brainliest to correct answer!
Artyom0805 [142]

Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.

\angle \: ABP =  \frac{180 - 80}{2}  = 50 \degree

Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.

Therefore, the angle BKQ is equal to 180-50-45=85°.

Of course angle BKP=180-85=95°.

Hope this helps :)

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2 years ago
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