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ELEN [110]
3 years ago
15

Charlotte wants to put a copy of a drawing on her website. She wants the copy of the drawing to be similar in shape but reduced

in size, as shown below.
Physical Drawing:
Height: 22.5 cm
Width: 7.5 cm

Copy:
Height: h
Width: 3 cm

What should the value of "h", the height in centimeters of the copy.

A) 2.5 cm
B) 7.5 cm
C) 9.0 cm
D) 15.0 cm
Mathematics
2 answers:
tekilochka [14]3 years ago
8 0
To find this you must find how much it has been reduced by so divide 7.5/3 which equals 2.5 if they are reduced by the same amount divide the other side by 2.5.


22.5/2.5=9 so the height of the copied picture should be 9 cm tall

C is your answer
alina1380 [7]3 years ago
6 0

This is talking about "similar" figures.
Two drawings are "similar" if they have the same shape but different size.

All squares are similar to all other squares.
All circles are similar to all other circles.
All equilateral triangles are similar to all other equilateral triangles.

For any other shapes (polygons that are not regular), you have to
check it out, and you have to know what you're looking for.

Here's the most important tool.  You should memorize it:

           If two figures are similar, then their corresponding sides
           are all in the same ratio.

That means that if I have two 4-sided shapes, and one is big and
one is small but they look like the same shape, then I need to look
at side BIG-A and side little-a and compare their lengths.
They won't be the same.  I can clearly see that one is big and
the other one is small.  What I need to do is divide the big one
by the little one.  That gives me their ratio. 

-- Let's say I divide their lengths and I get 2 .  (That means that
side BIG-A is double the length of side little-a.) 

OK.  I have the ratio of ONE pair of sides, a big one and a little one.

-- Now ... I pick another pair of sides:  side BIG-B and side little-b .
I divide those, and find their ratio.

-- Then ... I pick another pair of sides:  side BIG-C and side little-c .
I divide those, and find their ratio.

-- Finally ... I pick the last pair of sides:  side BIG-D and side little-d .

If I find that ALL of the ratios are the same number ... every side of the
BIG figure is 2 times (double) the length of the same side of the little
figure ... THEN the two figures are 'similar'.  They're really the same shape
but just different size.

Again, the Rule: 
In order to be similar, the corresponding pairs of all sides
in the figures must all be in the same ratio.

Fine.  Now you can solve Charlotte's problem.
I told you all that to tell you this:

Charlotte has two rectangles.  She wants them to be similar.
That tells us that (little length)/(BIG LENGTH) and (little width)/(BIG WIDTH)
must be the same ratio.

Great.  We know the ratio of their widths.  It's  3 / 7.5 =  0.4 .
Every dimension of the little rectangle has to be 0.4 of the
same dimension of the BIG one.

So she has to make the little length 0.4 of the BIG LENGTH .

The BIG LENGTH is  22.5 .
The little length has to be  (0.4 x 22.5) = 9 cm .

If you're confused, then stop here.
If you're still with me, I'll show you another thing about 'similar' figures:

Up to now, we only looked at the ratio of a BIG SIDE to a little side,
and we said they have to be the same number.

It turns out that IF the figures are similar, then the corresponding ratios
of two parts in the same figure are also equal.  What does that mean ?

I mean like the ratio of (BIG LENGTH)/(BIG WIDTH) and the ratio
of (little length)/(little width).

Look at the two rectangles Charlotte has now:

Length:  22.5
Width:    7.5        ratio of length/width
                          in the same rectangle = 3

Length:  9
Width:  3           ratio of length/width
                         in the same rectangle = 3          yay !

That also shows that the two rectangles are 'similar'.
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Answer:

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I roll a fair die twice and obtain two numbers X1= result of the first roll and X2= result of the second roll. Given that I know
azamat

By definition of conditional probability,

P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{P((X_1=4\text{ or }X_2=4)\text{ and }X_1+X_2=7)}{P(X_1+X_2=7)}

=\dfrac{P((X_1=4\text{ and }X_1+X_2=7)\text{ or }(X_2=4\text{ and }X_1+X_2=7))}{P(X_1+X_2=7)}

Assuming a standard 6-sided fair die,

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  • if X_2=4, then X_1=3.

Both outcomes are mutually exclusive with probability \frac1{36} each, hence total probability \frac2{36}=\frac1{18}.

Of the 36 possible outcomes, there are 6 ways to sum the integers 1-6 to get 7:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

and so a sum of 7 occurs \frac6{36}=\frac16 of the time.

Then the probability we want is

P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{\frac1{18}}{\frac16}=\frac13

6 0
3 years ago
A college student takes the same number of credits each semester. They had 5 credits when they started, and after 2 semesters, t
gogolik [260]

The rate at which the college student earns credit is 13 credits per semester.

<h3>What is an equation?</h3>

An equation is the statement that illustrates the variables given. In this case, two or more components are taken into consideration to describe the scenario.

Let the credit earned each semester be x.

This will be illustrated as:

5 + 2x = 31

Collect like terms

2x = 31 - 5

2x = 26

Divide

x = 26/2

x = 13

The rate is 13 credits per semester.

Learn more about equations on:

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5 0
1 year ago
Help me solve for x and y
Lena [83]
Let's start his with some trigonometry
Cos60=x/30
X=15
Now we can use Pythagorean therom
15^2+b^2=30^2
B(y)= 25.98
8 0
3 years ago
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Jenny made her uncle a quilt. The width is 7ft and the length is 5 ft. What is the area of the quilt? ​
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Answer:

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Step-by-step explanation:

A = wl

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A = 35

7 0
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