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Finger [1]
4 years ago
15

You originally draw a design for an art contest on a 2 inch by 5 inch card. The second phrase of the contest requires the drawin

g to be transferred to an 8.5 inch by 11 inch standard sheet of paper and utilize as much of the space on the paper as possible. You determine the largest size one of the dimensions of your drawing can be is 10.5 inches. What is the length of the other dimension if two drawings are similar? Type your exact answer in the blank without the units and round to the nearest tenth.
Mathematics
1 answer:
Bingel [31]4 years ago
4 0
4.2.
If the 2 inch side was scaled to 10.5 inches the 5 inch side would be scaled by 10.5/2 = 5.25, so would be 25.125 inches which is too big.

So the 10.5 inches is the 5 inch side on the original, and so it 10.5/5 = 2.1 times bigger.

Therefore the smaller side would be 2*2.1/= 4.2 inches.
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soldier1979 [14.2K]
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3 years ago
Triangle abc is equal to triangle pqr, angle p=8x+4, angle a=5x+16 , and angle q=10x+2. find angle r
saul85 [17]

Answer: b.not enough info

Step-by-step explanation:

Corresponding angles of congruent triangles are congruent, so \angle A =\angle P

However, we don't have all 3 interior angles of either triangle, so we cannot conclude anything.

8 0
2 years ago
The area of a regular octagon is 35 cm^2. What is the area of a regular octagon with sides five times as long?
Marat540 [252]
So... let's say the smaller regular octagon has sides of "x" long, then the larger octagon will have sides of 5x.

\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\

\bf \cfrac{small}{large}\quad \stackrel{area~ratio}{\cfrac{s^2}{s^2}}\implies \stackrel{area~ratio}{\cfrac{x^2}{(5x)^2}}\implies \stackrel{area~ratio}{\cfrac{x^2}{5^2x^2}}\implies \stackrel{area~ratio}{\cfrac{\underline{x^2}}{25\underline{x^2}}}=\stackrel{area~ratio}{\cfrac{35}{a}}
\\\\\\
\cfrac{1}{25}=\cfrac{35}{a}\implies a=\cfrac{25\cdot 35}{1}
3 0
4 years ago
Simplify: 6 + √27 / 3
Ray Of Light [21]

Answer:

6 + √27 / 3

6+ (3√3/3)

6+√3

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A rectangular parking area measuring 5000 ft squared is to be enclosed on three sides using​ chain-link fencing that costs ​$5.5
kaheart [24]

Answer:

Dimensions: 75.3778 ft and 66.3325 ft

Minimum price: $1658.31

Step-by-step explanation:

Let's call the length of the parking area 'x', and the width 'y'.

Then, we can write the following equations:

-> Area of the park:

x * y = 5000

-> Price of the fences:

P = 2*x*5.5 + y*5.5 + y*7

P = 11*x + 12.5*y

From the first equation, we have that y = 5000/x

Using this value in the equation for P, we have:

P = 11*x + 12.5*5000/x = 11*x + 62500/x

To find the minimum of this function, we need to take its derivative and then make it equal to zero:

dP/dx = 11 - 62500/x^2 = 0

x^2 = 65000/11

x = 250/sqrt(11) = 75.3778 ft

This is the x value that gives the minimum cost.

Now, finding y and P, we have:

x*y = 5000

y = 5000/75.3778 = 66.3325

P = 11*x + 62500/x = $1658.31

5 0
4 years ago
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