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ziro4ka [17]
3 years ago
13

If 30 Coins In Quarters And Dimes Are Worth $4.80, How Many Of Each Coin Are There?​

Mathematics
1 answer:
GREYUIT [131]3 years ago
3 0

Answer:

There are 18 dimes and 12 quarters

Step-by-step explanation:

Remember that

1 Dime =$0.10

1  Quarter=$0.25

Let

x-----> the number of dimes

y-----> the number of quarters

we know that

x+y=30

x=30-y ------> equation A

0.10x+0.25y=4.80

Multiply by 100 both sides

10x+25y=480 ------> equation B

Substitute equation A in equation B and solve for y

10(30-y)+25y=480

300-10y+25y=480

15y=180

y=12 quarters

Find the value of x

x=30-y=30-12=18 dimes

therefore

There are 18 dimes and 12 quarters

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The discriminant for the equation f(x)=6x^(2)-x+c is 73. What is the value of c?
Andrews [41]

Answer:

c = -3

Step-by-step explanation:

The discriminant  is 73, that means we have:

(-1)^2-4*6*c=73

then: 1-24c=73

or -24c= 73 -1

-24c = 72

Then c = 72/-24= -3

The answer is -3

Hope that useful for you.

5 0
3 years ago
Please someone help me...​
laiz [17]

Step-by-step explanation:

First factor out the negative sign from the expression and reorder the terms

That's

\frac{1}{ - (( \tan(2A) -  \tan(6A)  )}  -  \frac{1}{ \cot(6A)  -  \cot(2A) }

<u>Using trigonometric </u><u>identities</u>

That's

<h3>\cot(x)  =  \frac{1}{ \tan(x) }</h3>

<u>Rewrite the expression</u>

That's

\frac{1}{ - (( \tan(2A) -  \tan(6A)  )} -    \frac{1}{ \frac{1}{ \tan(6A) } }  -  \frac{1}{ \frac{1}{ \tan(2A) } }

We have

<h3>-  \frac{1}{  \tan(2A) -  \tan(6A)  } -   \frac{1}{ \frac{ \tan(2A) -  \tan(6A)  }{ \tan(6A) \tan(2A)  } }</h3>

<u>Rewrite the second fraction</u>

That's

<h3>-  \frac{1}{  \tan(2A) -  \tan(6A)  } -   \frac{ \tan(6A)  \tan(2A) }{ \tan(2A) -  \tan(6A)  }</h3>

Since they have the same denominator we can write the fraction as

-  \frac{1 +  \tan(6A) \tan(2A)  }{ \tan(2A) -  \tan(6A)  }

Using the identity

<h3>\frac{x}{y}  =  \frac{1}{ \frac{y}{x} }</h3>

<u>Rewrite the expression</u>

We have

<h3>-  \frac{1}{ \frac{ \tan(2A)  -  \tan(6A) }{1 +  \tan(6A) \tan(2A)  } }</h3>

<u>Using the trigonometric identity</u>

<h3>\frac{ \tan(x) -  \tan(y)  }{1 +  \tan(x)  \tan(y) }  =  \tan(x - y)</h3>

<u>Rewrite the expression</u>

That's

<h3>- \frac{1}{ \tan(2A -6A) }</h3>

Which is

<h3>-  \frac{1}{ \tan( - 4A) }</h3>

<u>Using the trigonometric identity</u>

<h3>\frac{1}{ \tan(x) }  =  \cot(x)</h3>

Rewrite the expression

That's

<h3>-  \cot( - 4A)</h3>

<u>Simplify the expression using symmetry of trigonometric functions</u>

That's

<h3>- ( -  \cot(4A) )</h3>

<u>Remove the parenthesis </u>

We have the final answer as

<h2>\cot(4A)</h2>

As proven

Hope this helps you

6 0
3 years ago
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Vladimir [108]
You're answers are correcy
3 0
3 years ago
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Marianna [84]

Answer:

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Now we have a visual so it'll be easier.

We know that this problem is a right triangle and we know that to solve for a missing side of a right triangle you use pythagorean theorem: a^2+b^2=c^2

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