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Deffense [45]
3 years ago
11

The perimeter of this rectangle is 24cm. If the length is 8 cm, what is the height?

Mathematics
2 answers:
gulaghasi [49]3 years ago
7 0

Answer:

h = 4 cm

Step-by-step explanation:

2(l + h) = 24 \\ l + h =  \frac{24}{2}  \\ l + h = 12 \\ 8 + h = 12 \\ ..(plug \: l = 8) \\ h = 12 - 8 \\ h = 4 \: cm \\

Gemiola [76]3 years ago
3 0

Answer:

4

Step-by-step explanation:

P = s+s+s+s

length = 8

8x2=16

24-16=8

8 / 2 = 4

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6 0
3 years ago
What is the answer to this one?
Kamila [148]

Answer:

20

Step-by-step explanation:

n-50=30

n=-30+50

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3 0
3 years ago
Manuel is riding his bike. The graph represents the distance Manuel travels from his house over time. Describe what is happening
olasank [31]

Answer:

Manuel is moving with a constant speed and reach his destination at 5 units distance from his house after 3 units of time.

He took rest at his destination for 4 units of time.

Then he reach his house at 5 units distance in 1 unit time.

Step-by-step explanation:

See the graph attached.

Manuel is riding his bike.

As the graph shows that Manuel starts his journey from his house and the time counts starts and then he is moving at a constant speed and reach his destination at 5 units distance from his house after 3 units of time.

The constant speed is \frac{5}{3} units.

Then he took rest at his destination for 4 units of time.

After that, he again starts to move towards his house at a constant speed and reach his house at 5 units distance in 1 unit time.

Therefore, his speed of return is \frac{5}{1} units. (Answer)

7 0
3 years ago
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NikAS [45]
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7 0
4 years ago
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An object is launched at 29.4 meters per
Lerok [7]

Answer:

The reasonable domain for the scenario is option 'a';

a) [0, 7]  

Step-by-step explanation:

For the projectile motion of the object, we are given;

The speed at which the object is launched, v = 29.4 meters per second

The height of the platform from which the object is launched, h = 34.3 meter

The equation for the height of the object as a function of time 'x' is given as follows;

f(x) = -4.9·x² + 29.4·x + 34.3

The domain for the scenario, is given by the possible values of 'x' for the function, which is found as follows;

At the height from which the object is launched, x = 0, and f(x) = 34.3

At the ground level to which the object can drop, f(x) = 0

∴ f(x) = -4.9·x² + 29.4·x + 34.3 = 0

-4.9·x² + 29.4·x + 34.3 = 0

By the quadratic formula, we have;

x = (-29.4 ± √(29.4² - 4 × (-4.9) × 34.3))/(2 × (-4.9)

∴ x = -1, or 7

Given that time is a natural number, we have the reasonable domain for the scenario as the start time when the object is launched, t = 0 to the time the object reaches the ground, t = 7

Therefore, the reasonable domain for the scenario is; 0 ≤ x ≤ 7 or [0, 7].

4 0
3 years ago
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