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2 years ago

Using the least number of coins how many quarters, dimes, nickels, and pennies to make $1.74

Mathematics

1 answer:

SpyIntel [72]2 years ago

4 0

Let's use our largest amount (quarter) to see how many can fit evenly into 1.74.

1.74/0.25 = 6.96
The most quarters that can fit into $1.74 is 6.

---Multiply $0.25 by 6 = $1.50

---Subtract $1.50 from $1.74 = $0.24

---We need to use the rest of the coins to fill up $0.24.

Now let's use our second largest amount (dime) to see how many can fit evenly into 0.24.

0.24/0.10 = 2.4
The most dimes that can fit into $0.24 is 2.

---Multiply $0.10 by 2 = $0.20

---Subtract $0.20 from $0.24 = $0.04

No nickels can fit into $0.04, because they are worth $0.05.

We can use our least amount (pennies) to fill in the $0.04 remaining.

Your answer is 6 quarters, 2 dimes, and 4 pennies.

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10 different total scores

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2 years ago

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Step-by-step explanation:

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3 years ago

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5 4 3 2 1

Step-by-step explanation:

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3 years ago

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2 years ago

In triangle ΔABC, ∠C is a right angle and CD is the height to

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Answer:

$m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha$

Step-by-step explanation:

The triangles are drawn below.

CD is perpendicular to AB as CD is height to AB.

Therefore, angles $m\angle CDB=m\angle CDA=90$ °

So, triangles ΔCBD and ΔCAD are right angled triangles.

Now, from the right angled triangle ΔABC,

$m\angle A+m\angle B =90\\\alpha+m\angle B=90\\m\angle B=90-\alpha$

From ΔCBD,

$m\angle CBD$ is same as $m\angle B$ .

So, $m\angle CBD=90-\alpha$

$m\angle BCD+m\angle BDC =90\\m\angle BCD+90-\alpha=90\\m\angle BCD=\alpha$

Now, from ΔCAD,

$m\angle CAD$ is same as $m\angle A$

So, $m\angle CAD=\alpha$

$m\angle CAD+m\angle ACD =90\\\alpha+m\angle ACD=90\\m\angle ACD=90-\alpha$

Hence, the unknown angles of both the triangles are:

$m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha$

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2 years ago