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eduard
3 years ago
14

Find the surface area of the two following figures. Show all your work.

Mathematics
1 answer:
Misha Larkins [42]3 years ago
7 0

Answer:

First shape: 366 in^2

Second shape: 376 in^2

Pretty sure these are correct, hope this helps!

You might be interested in
When is the reciprocal of a fraction a whole number? Plz explain
Sati [7]

The reciprocal of a fraction is only a whole number when
the numerator of the original fraction is ' 1 ' or a factor of
the denominator (original fraction is not in simplest form).

Any other time, the reciprocal of the fraction is another fraction.
(It'll be an improper fraction, and can be written as a mixed number.)


The reciprocal of a fraction is just the same fraction
turned upside down.

-- The reciprocal of  1/5  is  5/1  or just  5 .
   (Original numerator is ' 1 '.)

-- The reciprocal of  5/25  is  25/5  or just  5 .
    (Original fraction is not in simplest form.)

--  The reciprocal of  5/7  is  7/5 , or  1 and 2/5 . 

3 0
3 years ago
A girl that is 4 feet tall is standing next to the empire State building in New York city. The girls shadow is 3.2 feet long. If
lesya692 [45]

Answer: 1163.2 feet

<u>Step-by-step explanation:</u>

Set up the proportion then cross multiply:

\frac{height(girl)}{height(building)} = \frac{shadow (girl)}{shadow(building)}

\frac{4}{1454} = \frac{shadow (3.2)}{x}

4(x) = 1454(3.2)

   x = \frac{1454(3.2)}{4}

   x = 1163.2

3 0
3 years ago
Can someone help me?
horsena [70]
The answer is 16. ....
7 0
3 years ago
Solve the system of equations y = 40x2 and y = 19x + 3 algebraically.
ikadub [295]

We have been given a system of nonlinear equations.

\\y=40x^{2},y=19x+3

In order to solve this system we can first equate the two equations in order to get a quadratic in x.

\\40x^{2}=19x+3

Our next step is to bring all the terms on one side.

\\40x^{2}-19x-3=0

Now we have to solve this equation. We can solve it by factoring using the splitting middle term method.

\\40x^{2}-24x+5x-3=0

\\8x(5x-3)+(5x-3)=0

\\(8x+1)(5x-3)=0

Upon setting each of these factors equal to zero using zero product property we get

\\8x+1=0 \text{ or } 5x-3 = 0

Upon solving both these equations we get

\\x=-\frac{1}{8} \text{ or } x=\frac{3}{5}

Now we can substitute these values of x in the equation

y=19x+3

We get

\text{For }x=- \frac{1}{8} \Rightarrow y=19(- \frac{1}{8})+3= \frac{5}{8}\\

\text{For }x= \frac{3}{5}\Rightarrow y=19( \frac{3}{5})+3= \frac{72}{5}\\

Therefore, our final set of solutions are

(x,y)=(- \frac{1}{8},\frac{5}{8}) \text{ and } ( \frac{3}{5}, \frac{72}{5})


6 0
3 years ago
Read 2 more answers
PLEASE PLEASE PLEASE HELP
posledela
Not enough information to help with.
5 0
3 years ago
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