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3 years ago

Simplify the expression by combining like terms: 4n+4(2n+3)

2 answers:

8 0

First, you have to distribute 4(2n+3)
You then get 4n+8n+12
Secondly, you have to add 4n and 8n because they are like terms, so you get 12n
Lastly, the outcome is 12n+12 because 12n and 12 are not like terms.

Answer: 12n+12

Anna007 [38]3 years ago

7 0

First, you do the distributive property on:

4(2n+3)

You would get

8n+12

Then add the original part of the equation in (4n)

4n+8n+12

Combine like terms

12n+12 is the answer

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How do I evaluate the expression? What steps do I take?

Bingel [31]

Answer:

$\frac{4+8-2}{6-4}*3-1=\frac{12-2}{2}*3-1\\\\=\frac{10}{2}*3-1=5*3-1=15-1=14$

Step-by-step explanation:

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3 years ago

If P is parentheses and M is Multiplication and S is subtraction, what is EDA In PEMDAS? Very confused!

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E is exponents, D is division, A is addition

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3 years ago

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Two of the corresponding angles and the included sides of two triangles are congruent.Which statement must be true regarding the

lys-0071 [83]

Answer:

1. The two triangles are ASA congruent

2. The two triangles are SAS congruent

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2 years ago

You mow the lawn to earn your allowance. Each time you mow the lawn you earn $12 Write an equation for the number of dollars, d,

andrezito [222]

The equation is d=12m for amount you earn is 12 dollars per mow sesion.

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3 years ago

An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object

Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

$1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2 + \frac{1}{2} at^2\\3) v = v_0 + at$

What do we already know?

$v_0 = 29.4 ms^-1$
The launch is straight up
$a = -9.81 ms^-2$ this is the gravitational acceleration $g$

$s_0 = 0 m$ , since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, $v = 0 ms^-2$

$v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m$

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

$s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2$

this is the equation of the whole trajectory that object makes.

and if you solve this by making $s = 0$ , you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

5 0

3 years ago