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aniked [119]
2 years ago
9

You have $10 in your coin bank. Nickels (n) are worth $0.05, dimes (d) are worth $0.10, and quarters (q) are worth $0.25, each.

Which equation represents this situation?
Mathematics
2 answers:
kaheart [24]2 years ago
8 0
.05(n)+.10(d)+.25(q)=10
enyata [817]2 years ago
6 0

0.05(n) + 0.10(d) + 0.25(q) =10


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It is recommended that adults consume at least 1,000 mg of calcium every day. One ounce of whole milk contains about 25 mg of ca
gregori [183]

Given:

It is recommended that adults consume at least 1,000 mg of calcium every day.

25x+200y\geq 1000

Step-by-step explanation:

Let, x be the quantity of milk in ounces and y be the quantity of cheddar cheese in ounces.

A person meets the recommendation, if

25x+200y\geq 1000

For option A,

Substitute x=50 and y=2 in the given inequality.

25(50)+200(2)\geq 1000

1250+400\geq 1000

1650\geq 1000

This statement is true. So, option A is correct.

Similarly,

For option B,

Substitute x=0 and y=4.5 in the given inequality.

25(0)+200(4.5)\geq 1000

900\geq 1000

This statement is false. So, option B is incorrect.

For option C,

Substitute x=10 and y=3 in the given inequality.

25(10)+200(3)\geq 1000

850\geq 1000

This statement is false. So, option C is incorrect.

For option D,

Substitute x=39 and y=0 in the given inequality.

25(39)+200(0)\geq 1000

975\geq 1000

This statement is false. So, option D is incorrect.

Therefore, the correct option is only A.

3 0
2 years ago
Arnold took out a loan of 195,000 to purchase a home. At a 4.3% interest rate compounded annually, how much total will they have
kolbaska11 [484]

Answer:

25155000

Step-by-step explanation:

<h3>we know intrest (i):-PTR/100</h3>

by formula

  1. <u>195000×30×4.3%</u>

100

2. 5850000<u>×4.3</u>

<u>100</u>

100

3. 25155000

3 0
2 years ago
Lynn is making custom bricks. She combines 39 pounds of water and 48 pounds of brick dust in a mixing bucket. She spills 13 poun
ehidna [41]

Lynn mixes 39 pounds of water with 48 pounds of brick dust, so the mixture has a total weight of 39 + 48 = 87 pounds.

13 pounds are spilled during mixing, so she's left with 87 - 13 = 74 pounds to turn into bricks.

7 pounds are needed for 1 brick, so she can make 10 bricks because 7 • 10 = 70 and 7 • 11 = 77, and 70 < 74 < 77.

If she uses the mixture to make 10 bricks, she uses up 70 pounds of it, which leaves 4 pounds to be washed out.

5 0
2 years ago
What’s the answer? Because am having a hard time with this math problem.
Studentka2010 [4]

9514 1404 393

Answer:

  34.5 square meters

Step-by-step explanation:

We assume you want to find the area of the shaded region. (The actual question is not visible here.)

The area of the triangle (including the rectangle) is given by the formula ...

  A = 1/2bh

The figure shows the base of the triangle is 11 m, and the height is 1+5+3 = 9 m. So, the triangle area is ...

  A = (1/2)(11 m)(9 m) = 49.5 m^2

The rectangle area is the product of its length and width:

  A = LW

The figure shows the rectangle is 5 m high and 3 m wide, so its area is ...

  A = (5 m)(3 m) = 15 m^2

The shaded area is the difference between the triangle area and the rectangle area:

  shaded area = 49.5 m^2 - 15 m^2 = 34.5 m^2

The shaded region has an area of 34.5 square meters.

7 0
1 year ago
There are 11 portable mini suites (a.k.a. cages) in a row at the Paws and Claws Holiday Pet Resort. They are neatly labeled with
BigorU [14]

Step-by-step explanation:

a) 7!

If there are no restrictions, answer is 7! as it is the permutation of all animals.

b) 4! x 3!

As cats are 6 and Dogs are 5, thus 1st and last must be cats in order to have alternate arrangements. Therefore the only choices are the order of the cats among  themselves and the order of the dogs among themselves. There are 4! permutations of the cats and 3! permutations of the dogs, so there are a total of 4! x 3! possible arrangements of the suites.

c) 3! x 5!

There are 3! possible arrangements of  the dogs among themselves. Now, if we consider the dogs as  one ”object” together, then we can think of arranging the 4 cats  together with this 1 additional object. There are 5! such arrangements possible, so there are a total of 3! · 5! possible arrangements of the suites.

d) 2 x 4! x 3!

As required that all the cats must be together and all the dogs must be together, either the cats are all  before the dogs or the dogs are all before the cats. There are two possible arrangements thus two times of both possibilities is the answer i.e.  2 x 4! x 3!

3 0
2 years ago
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