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Gnesinka [82]
3 years ago
13

The areas of two similar squares are 16m and 49m.

Mathematics
2 answers:
Alexxandr [17]3 years ago
7 0

xy. 49.

uv. 16. the answer is 3.0625

OlgaM077 [116]3 years ago
7 0

Answer:

The scale factor of their side lengths is 4:7.

Step-by-step explanation:

Let the side length of two squares are p and q.

The area of a square is

A=a^2

Using this formula, we get the area of both squares.

A_1=p^2

A_2=q^2

It is given that the areas of two similar squares are 16m and 49m.

\frac{p^2}{q^2}=\frac{16}{49}

(\frac{p}{q})^2=\frac{16}{49}

Taking square root both the sides.

\frac{p}{q}=\sqrt{\frac{16}{49}}

\frac{p}{q}=\frac{4}{7}

Therefore the scale factor of their side lengths is 4:7.

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HELPPP ME PLEASE I REALLY NEED HELP
aliya0001 [1]
<h3>Answer:   choice F) 12 & 1/2 square yards.</h3>

=====================================================

We'll use this formula

a & b/c = (a*c + b)/c

to convert from a mixed number to an improper fraction.

Let's convert the first mixed number to an improper fraction

a & b/c = (a*c + b)/c

3 & 3/4 = (3*4 + 3)/4

3 & 3/4 = 15/4

Repeat the same idea for the second mixed number as well

a & b/c = (a*c + b)/c

3 & 1/3 = (3*3 + 1)/3

3 & 1/3 = 10/3

--------------------

In short we have

  • 3 & 3/4 = 15/4
  • 3 & 1/3 = 10/3

Doing these conversions is useful to be able to multiply the values like so:

(15/4)*(10/3) = (15*10)/(4*3) = 150/12 = (6*25)/(6*2) = 25/2

Now use your calculator to find that 25/2 = 12.5

We can then say....

12.5 = 12 + 0.5 = 12 + (1/2) = 12 & 1/2

The answer is choice F) 12 & 1/2 square yards.

6 0
2 years ago
what is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube
Dmitrij [34]
Let the least possible value of the smallest of 99 cosecutive integers be x and let the number whose cube is the sum be p, then

\frac{99}{2} (2x+98)=p^3 \\  \\ 99x+4,851=p^3\\ \\ \Rightarrow x=\frac{p^3-4,851}{99}

By substitution, we have that p=33 and x=314.

Therefore, <span>the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube is 314.</span>
3 0
3 years ago
Please help i dont get this at all
Tom [10]

Answer:

\frac{20y^{2} a^{2} }{41bx^{3} }

Step-by-step explanation:

1) Flip the second fraction and change the operation to multiplication:

\frac{5x^{2}y^{3}  }{2a^{5}b^{4}  } *  \frac{8a^{7}b^{3}  }{41x^{5}y  }

2) Cross cancel factors:

\frac{5y^{2}   }{2b  } *  \frac{8a^{2}  }{41x^{3}  }

3) Multiply:

\frac{40y^{2} a^{2} }{82bx^{3} }

4) Simplify:

\frac{20y^{2} a^{2} }{41bx^{3} }

4 0
3 years ago
Read 2 more answers
Round to the nearest tenth if necessary.
mihalych1998 [28]

Distance formula: sqrt((x2-x1)^2 + (y2-y1)^2)


distance = sqrt((-1 -2)^2 + (-5-5)^2)
distance = sqrt(109)

Distance = 10.44

3 0
2 years ago
HELP ASAP!!!
IrinaVladis [17]

Answer:

0.4, 0.8, 1.2 ,1.6 ,2.0

Step-by-step explanation:

First we have to find what the question is really asking and get rid of unwanted or not needed information. It is asking to write a numeric sequence for the first 5 minutes of the 30 minutes.

Second we need to find the pattern. So lets divide $12 by 30. That will equal 0.4 a minute.

Last we add 0.4 for each minute. 0.4 is the first, then 0.8 is 2 minutes, then 1.2 and etc.

7 0
2 years ago
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