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Katena32 [7]
3 years ago
12

each of five friends has X action figures in his or her collection. Each friend buys 11 more action figures. Now the five friend

s have a total of 120 action figures.write an equation that models this problem then solve
Mathematics
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

Each friend has 13 actions

Step-by-step explanation:

Let x represent the actions of each friends

5friends will have 5x actions

And each friend has 11 more actions which means 5(11)=55

It means 5x+55

The total of their actions is 120

It can be written as 5x+55= 120

To solve

5x+55=120

Substrate 55 from both sides

5x=120-55

5x=65

X=65/5

x=13

Which each friend has 13 actions

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I have no clue what the answer is.
Vera_Pavlovna [14]

Area of the word ME = 20 square units

Solution:

Picks' Theorem:

$Area=I+\frac{B}{2}-1$

where <em>I </em>is the interior points and <em>B</em> is the points on the boundary.

Let us first find the area of the letter M:

Number of interior points in M = 0

Number of points in the boundary = 22

Using pick's theorem,

Area of M = 0+\frac{22}{2}-1

Area of M = 10 square units

Now, find the area of the letter E:

Number of interior points in E = 0

Number of points in the boundary = 22

Using pick's theorem,

Area of E = 0+\frac{22}{2}-1

Area of E = 10 square units

Area of the word ME = Area of M + Area of E

                                   = 10 square units + 10 square units

Area of the word ME = 20 square units

7 0
4 years ago
For a moving object, the force acting on the object varles directly with the object's acceleration. When a force of 25 N acts on
Radda [10]

\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}

According to the Question,

\large{\sf{ f \propto \: a}}

<h3>From the Question,</h3>

  • Initial Acceleration,a = 5 m/s²

  • Force at a = 5 m/s²,f = 25 N

  • Final Acceleration,A = 9 m/s²

<h3>To find</h3>

Force exerted on the object at a = 9 m/s²

We Know that,

\huge{\boxed{ \boxed{ \sf{f = ma}}}}

Firstly,we need to find the mass of the object

Thus,

\sf{m =  \frac{f}{a} } \\  \\  \implies \:  \sf{m =  \frac{25}{5} } \\  \\  \huge{ \implies \:  \sf{m = 5 \: kg}}

Now,

\sf{F = mA} \\  \\  \implies \:  \sf{F = (5)(9)} \\  \\  \huge{ \implies \:  \sf{F = 45 \: N}}

7 0
3 years ago
Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Which shows the correct substitution of the
Brums [2.3K]
A = 1
b = -2
c = -3

And when you plug in and solve you'll get correct answers of x = 3 and x = -1
5 0
3 years ago
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Write an expression for the sequence of operations described below.
xxTIMURxx [149]

2( \frac{ {5}^{3} }{b})
5 0
3 years ago
Omg please help me!!!
marin [14]

area = pi * r^2

here, the radius is 2 as its the diameter/2.

so the area is 4pi.

circumference = 2*pi*r.

so the circumference is 4pi as well.

4 0
2 years ago
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