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3 years ago

5

The dimensions of an office conference room are 15 feet by 27 feet. if the conference room blueprint dimensions are 5 centimeter

s by 9 centimeters, what is the scale of the blueprint?

2 answers:

Tasya [4]3 years ago

8 0

The scale of the blueprint is 1cm= 3ft

marysya [2.9K]3 years ago

6 0

Answer:

1cm=4feet

on study island

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The diameter of a circular tablecloth is 72 inches. Find the area and circumference. Use 3.14 as an approximation for pi.

nalin [4]

Answer:

Part 1) The area is $A=4,069.44\ in^{2}$

Part 2) The circumference is $C=226.08\ in$

Step-by-step explanation:

step 1

Find the area

we know that

The area of a circle (circular tablecloth) is equal to

$A=\pi r^{2}$

we have

$r=72/2=36\ in$ ------> the radius is half the diameter

$\pi =3.14$

substitute

$A=(3.14)(36)^{2}$

$A=4,069.44\ in^{2}$

step 2

Find the circumference

we know that

The circumference of a circle (circular tablecloth) is equal to

$C=\pi D$

we have

$D=72\ in$

$\pi =3.14$

substitute

$C=(3.14)(72)$

$C=226.08\ in$

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3 years ago

En el pequeño frasco se colocaron 7 galletas para regalar si ornearon 56 ¿cuantos frascos. regalaron?

Solnce55 [7]

Answer:

1 frasco

Step-by-step explanation:

En el pequeño frasco se colocaron 7 galletas para regalar si ornearon 56 ¿cuantos frascos. regalaron?

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2 years ago

Given the piece wise function shown below, select all of the statements that are true

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C and d

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and
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3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

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GREYUIT [131]

Answer:

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Step-by-step explanation:

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3 years ago

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