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3 years ago

10

A photo 5 inches wide and 8 inches long is enlarged. How long is the new photo if it is (1)10 inches; (2)15 inches; (3)8 inches

wide?

2 answers:

sweet [91]3 years ago

5 0

1. 16 inches long, 2. 24 inches long, 3. 12.8 inches long

telo118 [61]3 years ago

4 0

Answer:

1.16 in

2.24 in

3.12.8 in

Step-by-step explanation:

We are given that

Width of photo=5 in

Length of photo=8 in

We have to find the length of new photo in each case.

1. Let x represent the width and y represents the length of photo.

When width increase then length of photo is also increases then it is direct proportion.

$\frac{x_1}{y_1}=\frac{x_2}{y_2}$

Substitute the values then we get

$\frac{5}{8}=\frac{10}{y}$

$y=\frac{10\times 8}{5}=16$

Hence, the length of photo=16 in

2.Width=x=15 in

Again using the formula

$\frac{5}{8}=\frac{15}{y}$

$y=\frac{15\times 8}{5}=24$

Hence, the length of new photo=24 in

3. Width =x=8 in

Substitute the values in the given formula

$\frac{5}{8}=\frac{8}{y}$

$y=\frac{8\times 8}{5}=\frac{64}{5}=12.8$

Hence, the length of new photo=12.8 in

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arsen [322]

Answer:

x=1

Step-by-step explanation:

We have,

h(x) = 5-2x

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Now,

h(x)=5-2x=3

or,5-2x=3

or,5-3=2x

or,2/2=x

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Valentin [98]

<em>The question is not clearly readable, but I'm assuming the expression which makes more sense, so you can have a clue</em>

Answer:

$i^{97}-i=0$

Step-by-step explanation:

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The imaginary unit i is defined as

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The first powers of i are

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Olenka [21]

Answer:

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Step-by-step explanation:

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$d_1=\dfrac{50}{tan (61^o)}$

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And the distance to the ship B is $d_2$ and is given by

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