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denpristay [2]
3 years ago
7

I will give brainliest if correct

Mathematics
1 answer:
m_a_m_a [10]3 years ago
4 0

Answer:

The surface area of the net shown is: 88 in^2

Step-by-step explanation:

Break up the different shapes in the net:

Let's take the big square first:

It has a side of 5in and a base of 4in.

We multiply both these values to get 20in^2

Lets look at the small rectangular box now:

It has a side of 3in and a base of 4in.

We multiply both these values to get 12in^2

There are two big squares so we add up their surface areas:

20in^2 + 20in^2 = 40in^2

There are 4 small rectangular squares so we add up their surface areas:

12in^2 + 12in^2 + 12in^2 + 12in^2 = 48in^2

Now we add up the big square and small rectangular squares values:

40in^2 + 48in^2 = 88in^2

Therefore the answer is 88in^2

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Answer: they are he same as 75%

Step-by-step explanation:

onvert 3/4 to a percent. Begin by converting the fraction 3/4 into decimal. Multiply the decimal by 100 and write the result with the percentage sign: 0.75 × 100 = 75%.

6 out of 8 can be written as 6/8 and equals to 75%. Let's understand the conversion of a fraction to a percentage. To find the percent for this fraction, we have to find the number of parts that would be shaded out of 100. To convert a fraction to percent, we multiply it by 100/100.

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The answer to this picture
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Step-by-step explanation:

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3 years ago
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Given the information in the diagram below, find the measure of angle N. Use RACE and complete sentences to show how you found t
Sonja [21]

Answer:

m<N = 76°

Step-by-step explanation:

Given:

∆JKL and ∆MNL are isosceles ∆ (isosceles ∆ has 2 equal sides).

m<J = 64° (given)

Required:

m<N

SOLUTION:

m<K = m<J (base angles of an isosceles ∆ are equal)

m<K = 64° (Substitution)

m<K + m<J + m<JLK = 180° (sum of ∆)

64° + 64° + m<JLK = 180° (substitution)

128° + m<JLK = 180°

subtract 128 from each side

m<JLK = 180° - 128°

m<JLK = 52°

In isosceles ∆MNL, m<MLN and <M are base angles of the ∆. Therefore, they are of equal measure.

Thus:

m<MLN = m<JKL (vertical angles are congruent)

m<MLN = 52°

m<M = m<MLN (base angles of isosceles ∆MNL)

m<M = 52° (substitution)

m<N + m<M° + m<MLN = 180° (Sum of ∆)

m<N + 52° + 52° = 180° (Substitution)

m<N + 104° = 180°

subtract 104 from each side

m<N = 180° - 104°

m<N = 76°

4 0
3 years ago
if h=-16t ²+160t represents the height of a rocket in T seconds after it was fired when will the rocket hit the ground?
makkiz [27]

Answer:

The time after which the rocket hit the ground is 5 seconds

Step-by-step explanation:

Given as :

The distance of the cover by rocket at the height h = - 16 t² + 160 t

Let the time after which rocket hit the ground = T seconds

Now, When the rocket hits the ground, then at that time, the velocity of the rocket becomes zero.

I.e velocity = \frac{\partial h}{\partial t} = 0

Or, v = \frac{\partial h}{\partial t} = 0

Now, v = \frac{\partial h}{\partial t}

Or, v = \frac{\partial (-16t^{2}+160t)}{\partial t}

or, v = - 32 t + 160

Now, ∵ velocity of rocket after reaching the ground becomes zero

So, v = - 32 t + 160 = 0

Or, 32 t = 160

Or, t = \frac{160}{32}

∴ t = 5 sec

So, The time after which the rocket hit the ground = 5 sec

Hence The time after which the rocket hit the ground is 5 seconds answer

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Solve the equation. 42=d^2-22
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