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2 years ago

6

A taxi cab charges $1.75 for the flat fee and $0.25 for each mile. Write an in equality to determine how many miles Eddie can tr

avel if he has $15 to spend.

1 answer:

Gnesinka [82]2 years ago

5 0

Answer:

.25x+1.75=15 and x=53

Step-by-step explanation:

first subtract 1.75 from both sides

then divide by .25 on both sides

and you will get x=53

You might be interested in

Write an expression for the sequence of operations described below.

SSSSS [86.1K]

(3-6)-p

It says your subtracting 6 from 3 which is (3-6) and it says "then" that's why we put parentheses and then put "-p"

5 0

2 years ago

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Lan scores 5 out of 60 mark test.what is his score as a percentage to 1 decimal place

artcher [175]

Answer:

.1 approx.

Step-by-step explanation:

5/60 and x/100%

Cross multiplying...

5 x 100% = 5

5/60 ~= .1

( I dunno if I am right, so please check first.)

6 0

3 years ago

Which values are solutions of the inequality 5 < y -8<br><br>8 , 9. , 13. , 15. , 20

Sladkaya [172]

Answer:

5 is greater than y-8 since 5 is positive it + in the begging +5 while - 8 is less than 0 because it has - in the begging

6 0

2 years ago

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

Learn more here: brainly.com/question/11689670

6 0

2 years ago

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The area of the scale model of a playground is 6 square yards. The scale model is enlarged by a scale factor of 3 to create the

REY [17]

Answer:

B: 54 square feet

Step-by-step explanation:

There are 3 feet in 1 yard. The scale model is 6 square yards.

Multiply 6 by 3 to get the amount of square yards.

There are 18 square yards in the actual playground.

Since there are 3 feet in a yard, multiply 18 by 3.

This gives you your answer:

54 square feet

6 0

3 years ago

Read 2 more answers