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swat32
3 years ago
5

A model of the is 2.9 ft tall and 1.2 ft wide. The Eiffel Tower is actually 410 ft wide. What is the actual height of the Eiffel

Tower?
Mathematics
2 answers:
natima [27]3 years ago
7 0
\bf \cfrac{height}{width}\qquad \stackrel{model}{\cfrac{2.9}{1.2}}\qquad \stackrel{actual}{\cfrac{x}{410}}\qquad \qquad \cfrac{2.9}{1.2}=\cfrac{x}{410}\implies \cfrac{2.9\cdot 410}{1.2}=x
const2013 [10]3 years ago
7 0

Answer:

<h2><em>its </em><u><em>B)990.8 ft</em></u></h2>

Step-by-step explanation:

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5. Sally took money from her bank account to go
Anit [1.1K]

Answer:

110

Step-by-step explanation:

8 0
2 years ago
Find the value of the variable.
nalin [4]

Answer:

The variable, y is 11°

Step-by-step explanation:

The given parameters are;

in triangle ΔABC;          {}              in triangle ΔFGH;

Segment \overline {AB} = 14         {}               Segment \overline {FG} = 14

Segment \overline {BC} = 27         {}              Segment \overline {GH} = 19

Segment \overline {AC} = 19         {}               Segment \overline {FH} = 2·y + 5

∡A = 32°                       {}                ∡G = 32°

∡A = ∠BAC which is the angle formed by segments \overline {AB} = 14 and \overline {AC} = 19

Therefore, segment \overline {BC} = 27, is the segment opposite to ∡A = 32°

Similarly, ∡G = ∠FGH which is the angle formed by segments \overline {FG} = 14 and \overline {GH} = 19

Therefore, segment \overline {FH} = 2·y + 5, is the segment opposite to ∡A = 32° and triangle ΔABC ≅ ΔFGH by Side-Angle-Side congruency rule which gives;

\overline {FH} ≅ \overline {BC} by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

∴ \overline {FH} = \overline {BC} = 27° y definition of congruency

\overline {FH} = 2·y + 5 = 27° by transitive property

∴ 2·y + 5 = 27°

2·y = 27° - 5° = 22°

y = 22°/2 = 11°

The variable, y = 11°

8 0
2 years ago
A person wants to create a vegetable garden and keeps the rabbits out by enclosing it with 100 feet of fencing. The area of the
alekssr [168]

Answer:

The garden can not have area of 700ft^2

Step-by-step explanation:

We are given equation for area as

A(w)=w\times (50-w)

where

w is the width (in feet) of the garden

now, we are given area =700 ft^2

so, we can set area =700

and then we can solve for w

w\times (50-w)=700

50w-w^2=700

-w^2+50w-700=0

now, we can use quadratic formula

ax^2+bx+c=0

w=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

we can compare and find a,b, and c

a=-1 , b=50 , c=-700

now, we can plug values

w=\frac{-50\pm \sqrt{50^2-4\left(-1\right)\left(-700\right)}}{2\left(-1\right)}

w=25-5\sqrt{3}i,\:w=25+5\sqrt{3}i

We can see that values of w is not real

so, w does not exists when area =700

so, The garden can not have area of 700ft^2


4 0
3 years ago
Demi and Juan start saving money for their vacation. Demi starts with $50 and saves $20 each week; Juan started with only $35, b
Fiesta28 [93]

Answer:

i d k

Step-by-step explanation:

7 0
2 years ago
Luis said that when his dog was born, she had a mass of 370 grams and that when fully grown, his dog had a mass of 37 kilograms.
yanalaym [24]
I do not agree. This is because when the dog was a puppy, she was 370 grams not kilograms. Diego's mistake was mistaking grams for kilograms.
4 0
3 years ago
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