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3 years ago

Write the next three terms of the geometric sequence.

Mathematics

1 answer:

Sati [7]3 years ago

8 0

Answer:

5, 1, 0.2

Step-by-step explanation:

Divide the number by 5 to get the next one.

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Can you help me with this problem please. I will mark brainlist

schepotkina [342]

Answer:

5.5 pt or 5 1/2 pt

Step-by-step explanation:

A pint is half of a quart, so multiply 2 3/4, or 2.75 by 2, and you'll get your answer. If

6 0

2 years ago

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Fran has a poster that measures 3/4 square yard in area the length of the poster is 1/3 yard what is the width of the poster

Harrizon [31]

Answer:

9/4 yards

Step-by-step explanation:

A poster is a rectangle.

Area of rectangle = Length × Width

Given:

Area = 3/4 yd

Length = 1/3 yd

Substitute the given values to the equation.

3/4 = 1/3 × w

3/4 = 1/3w

3/4 ÷ 1/3 = w

3/4 × 3/1 = w

9/4 = w

6 0

3 years ago

Japhus Moore's credit card statement showed a previous balance of $921.07 , a financial charge of $13.80, new purchases of $962.

lara [203]

Answer:

$906.51

Step-by-step explanation:

5 0

3 years ago

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There are four movies showing at the theater, 2/5 of the people bought tickets for the comedy, 1/4 bought tickets for the horror

RUDIKE [14]

Given:

Number of people bought tickets for comedy = $\frac{2}{5}$

Number of people bought tickets for horror = $\frac{1}{4}$

Number of people bought tickets for kids movie = $\frac{3}{10}$

To find the number of people who bought tickets for the action movie.

Let us take,

Total number of tickets = 1

Now,

Total number of tickets for comedy, horror and kids movie are = $(\frac{2}{5} +\frac{1}{4} +\frac{3}{10} )$

So,

$(\frac{2}{5} +\frac{1}{4} +\frac{3}{10} )$

LCM of 5,4,10 = 20

= $\frac{8+5+6}{20}$

= $\frac{19}{20}$

Rest are action movie's tickets.

Therefore,

Number of action movie tickets = $1-\frac{19}{20}$

= $\frac{20-19}{20}$

= $\frac{1}{20}$

Hence,

The number of people who bought the tickets for action movie is $\frac{1}{20}$ .

7 0

3 years ago

Find the expression that represents the length of the hypotenuse of a right triangle whose legs measure m^2-n^2 and 2mn

My name is Ann [436]

$\bf c^2=a^2+b^2\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
------\\
a=m^2-n^2\\
b=2mn
\end{cases}\\\\\\ c^2=(m^2-n^2)^2+(\underline{2mn})^2
\\\\\\
c^2=m^4-2m^2n^2+n^4+\underline{2^2m^2n^2}
\\\\\\
c^2=m^4-2m^2n^2+n^4+{4m^2n^2}
\\\\\\
c^2=m^4+2m^2n^2+n^4\impliedby \textit{perfect square trinomial}
\\\\\\
c^2=(m^2+n^2)^2\implies c=\sqrt{(m^2+n^2)^2}\implies c=m^2+n^2$

7 0

3 years ago