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3 years ago

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A student wants to reduce the size of a photograph that is 12 inches wide. She selects 75% on the photocopy machine. What will b

e the width of the copy?

2 answers:

LenKa [72]3 years ago

8 0

Answer:

Step-by-step explanation:

(12)*(0,75) = 9 inches

The width of the copy will be 9 inches

Best regards

yKpoI14uk [10]3 years ago

4 0

Answer:

9 inches

Step-by-step explanation:

The length and width of the photograph will become 75% of the original dimensions. The original width is 12 inches. The reduced with is 75% of 12 inches. To find a percent of a number, multiply the percent by the number. Change the percent to a decimal by dividing the percent by 100, which means move the decimal of the percent two places to the left.

75% of 12 inches =

= 0.75 * 12 inches

= 9 inches

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The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

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Point F is on circle C. What is the length of line segment GF?

Levart [38]

The answer:
the full question is as follow:
<span>Point F is on circle C. What is the length of line segment GF?

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15.0 units
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CF is the radius of the circle, so it is CF =CE = 7.5
all that we want to find is the value of GC
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that is, GE² + EC² = GC², and from this, we have GC = sqrt(</span>GE² + EC² )

GC = sqrt(10² + 7.5² )=sqrt(56.25)=12.5
<span>
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