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3 years ago

12

Paul owes PAula $.35 and has a pocket full of five cents $.10 coins and $.25 coins that he can use to pay her what is the differ

ence between the largest in the smallest number of coins he can use to pay her

1 answer:

coldgirl [10]3 years ago

4 0

He can use:
a quarter and a dime (difference: 15)
a quarter and two nickels (difference: 20)
a dime and three nickels (difference: 5)
two dimes and a nickel (difference: 5)

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Find the area 20 points!!!

Dmitriy789 [7]

Answer:

the answer would be 29 :)

Step-by-step explanation:

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3 years ago

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Let f(t) give the number of liters of fuel oil burned in t days, and w(r) the liters burned in r weeks. Find a formula for w by

viktelen [127]

Answer:

7 f(t)

Step-by-step explanation:

So, our f(t) is the number of liters burned in t days. If t is 1, f(t)=f(1) and so on for every t.

w(r) id the number of liters in r weeks. This is, in one week there are w(1) liters burned.

As in one week there are 7 days, we can replace the r, that is a week, by something that represents 7 days. As 1 day is represented by t, one week can be 7t (in other words r = 7t). So, we have that the liters burned in one week are:

w(r) = w[7f(t)]

So, we represented the liters in one week by it measure of days.

So, we can post that the number of liters burned in 7 days is the same as the number of liters burned 1 day multiplied by 7 times. So:

w (r) = w[7 f(t)] = 7 f(t)

Here we hace the w function represented in terms of t instead of r.

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3 years ago

In 2012, a health insurance plan for a 35-year-old and his or her spouse costs $301 per month. That rate increased to $430 p

katrin2010 [14]

Answer: See explanation

Step-by-step explanation:

Let the cost for insuring the applicant = a.

Let the cost for insuring the spouse = b

Let the cost for insuring the first child= c

Let the cost for insuring the second child = d

A 35-year-old health insurance plan and that of his or her spouse costs $301 per month. This means that:

a + b = $301.

That rate increased to $430 per month if a child were included. This means the cost of a child will be:

= $430 - $301

= $129

The rate increased to $538 per month if two children were included. This means the cost for the second child will be:

= $538 - $430

= $108

The rate dropped to $269 per month for just the applicant and one child. His will be the cost of the applicant and a single child. This can be written as:

a + $129 = $269

a = $269 - $129

a = $140

Since a + b = $301

$140 + b = $301

b = $301 - $140

b = $161

Applicant = $140

The spouse = $161

The first child = $129

The second child = $108

3 0

3 years ago

Help me please ;-;!!!!! Use a coordinate grid to create a map of a town with at least five different locations, such as a house,

Sholpan [36]

A. Our diagram have six location: my house (2,1), a movie theater (5,3), the school (6,6), the police station (3,-2), the airport (-3,3), and the mall (-2,-3)

B. Distance between my house and the movie theater:
To calculate the distance we are going to use the points (2,1) and (5,3) to create a right triangle with tow legs of measures 3 and 2; the hypotenuse of the right triangle will be the distance between my house and the movie theater.
Using the Pythagorean theorem:
$d^2=3^2+2^2$
$d^2=9+4$
$d^2=13$
$d= \sqrt{13}$
We can conclude that the distance between my house and the movie theater is $\sqrt{13}$

Our second distance is the distance between the police station and the airport:
This time we are using the points (3,-2) and (-3,3) to create a right triangle with legs of measures 6 and 5; the hypotenuse of our triangle will be the the distance between our points:
$d^2=6^2+5^2$
$d^2=36+25$
$d^2=61$
$d= \sqrt{61}$
We can conclude that the distance between the police station and the airport is $\sqrt{61}$

C. Distance form the park to the Fire Station:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance from the park to the Fire Station is $\sqrt{10}$

Distance from the stadium to the mall:
$d^2=2^2+1^2$
$d^2=4+1$
$d^2=5$
$d= \sqrt{5}$
We can conclude that the distance from the stadium to the mall is $\sqrt{5}$

D. What is the distance between your house and the mall?
Answer:
$d^2=4^2+4^2$
$d^2=16+16$
$d^2=32$
$d= \sqrt{32}$
$d=4 \sqrt{2}$
We can conclude that the distance between my house and the mall is $4 \sqrt{2}$

What is the distance between the movie theater and the school?
answer:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance between the movie theater and the school is $\sqrt{10}$

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3 years ago

An airplane is flying at an altitude of 33,000 feet. It needs to reduce its altitude by 6% to avoid some turbulence. After it pa

Tema [17]

Step-by-step explanation:

100% = 33,000 ft

1% = 100%/100 = 33000/100 = 330 ft

6% = 1% × 6 = 330 × 6 = 1980 ft

that means the plane has to go down 1980 ft to

33,000 - 1980 = 31,020 ft

and then it goes up by 5000 ft :

31,020 + 5000 = 36,020 ft

that is the final altitude.

5 0

2 years ago

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