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S_A_V [24]
3 years ago
14

Samuel has a collection of toy cars. his favorites are the 27 red ones witch make up 60%of his collection. how many toy cars doe

s Samuel have?????
who ever answers first i will put in as best answer...ready go!!!!!
Mathematics
2 answers:
Leokris [45]3 years ago
4 0

The <em>correct answer</em> is:


45 cars.


Explanation:


Let x represent the total number of cars in Samuel's collection.


We know that 27 is 60% of the total. We write 60% of the total as 0.6x. Since this is 27, this gives us the equation


0.6x = 27


Divide both sides by 0.6:

0.6x/0.6 = 27/0.6


x = 45

Genrish500 [490]3 years ago
3 0
Number of cars Samuel has = 27 * 100 / 60 = 45 Cars
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We assume the brightness of a dim day to be x.

According to the developed scale, the brightness of an illuminated day will be 4 times that of a dim day.

Thus, the brightness of an illuminated day = 4*the brightness of a dim day = 4x.

According to the developed scale, the brightness of a radiant day will be 4 times that of an illuminated day.

Thus, the brightness of a radiant day = 4*the brightness of an illuminated day = 4*4x = 16x.

Now, the ratio of the brightness of a radiant day to the brightness of a dim day = 16x:x = 16x/x = 16:1.

Thus, according to the developed scale, a radiant day is <u>16 times</u> brighter than a dim day.

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The question provided is incomplete. The complete question is:

"Suppose you have developed a scale that indicates the brightness of sunlight. Each category in the table is 4 times brighter than the next lower category. For example, a day that is dazzling is 4 times brighter than a day that is radiant. How many times brighter is a radiant day than a dim day?

Dim=2

Illuminated=3

Radiant=4

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Terminos semejantes de -7a²(4a-3)​
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3 years ago
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A pumpkin is thrown horizontally off of a building at a speed of 2.5\,\dfrac{\text m}{\text s}2.5 s m ​ 2, point, 5, start fract
4vir4ik [10]

Answer:−47.0

​

​

Step-by-step explanation:Step 1. List horizontal (xxx) and vertical (yyy) variables

xxx-direction yyy-direction

t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text

a_x=0a

x

​

=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a

y

​

=−9.8

s

2

m

​

a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction

\Delta x=12\,\text mΔx=12mdelta, x, equals, 12, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text

v_x=v_{0x}v

x

​

=v

0x

​

v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=\text ?v

y

​

=?v, start subscript, y, end subscript, equals, start text, question mark, end text

v_{0x}=2.5\,\dfrac{\text m}{\text s}v

0x

​

=2.5

s

m

​

v, start subscript, 0, x, end subscript, equals, 2, point, 5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v

0y

​

=0v, start subscript, 0, y, end subscript, equals, 0

Note that there is no horizontal acceleration, and the time is the same for the xxx- and yyy-directions.

Also, the pumpkin has no initial vertical velocity.

Our yyy-direction variable list has too many unknowns to solve for v_yv

y

​

v, start subscript, y, end subscript directly. Since both the yyy and xxx directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:

\Delta x=v_xtΔx=v

x

​

tdelta, x, equals, v, start subscript, x, end subscript, t

Once we know ttt, we can solve for v_yv

y

​

v, start subscript, y, end subscript using the kinematic equation that does not include the unknown variable \Delta yΔydelta, y:

v_y=v_{0y}+a_ytv

y

​

=v

0y

​

+a

y

​

tv, start subscript, y, end subscript, equals, v, start subscript, 0, y, end subscript, plus, a, start subscript, y, end subscript, t

Hint #22 / 4

Step 2. Find ttt from horizontal variables

\begin{aligned}\Delta x&=v_{0x}t \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ &=\dfrac{12\,\text m}{2.5\,\dfrac{\text m}{\text s}} \\\\ &=4.8\,\text s \end{aligned}

Δx

t

​

 

=v

0x

​

t

=

v

0x

​

Δx

​

=

2.5

s

m

​

12m

​

=4.8s

​

Hint #33 / 4

Step 3. Find v_yv

y

​

v, start subscript, y, end subscript using ttt

Using ttt to solve for v_yv

y

​

v, start subscript, y, end subscript gives:

\begin{aligned}v_y&=v_{0y}+a_yt \\\\ &=\cancel{0\,\dfrac{\text m}{\text s}}+\left(-9.8\,\dfrac{\text m}{\text s}\right)(4.8\,\text s) \\\\ &=-47.0\,\dfrac{\text m}{\text s} \end{aligned}

v

y

​

​

 

=v

0y

​

+a

y

​

t

=

0

s

m

​

​

+(−9.8

s

m​

)(4.8s)

=−47.0

s

m

5 0
2 years ago
Read 2 more answers
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