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Strike441 [17]
2 years ago
7

There are nickles and quarters worth $2.20 in total. If there are 28 coins, how many nickels are there?

Mathematics
1 answer:
Eduardwww [97]2 years ago
6 0
Answer:

There are 24 nickels

Step-by-step explanation:

Let x represent the number of nickels

Let y represent the number of quarters
—————————————————————

Value Value
Type Number of of
of of each all
Coin Coin Coin Coin
—————————————————————
Nickels | x | $0.05 | $0.05x
Quarters | y | $0.25 | $0.25y
—————————————————————
Totals 28 ——— $2.20
•••••••••••••••••••••••••••••••••••••••••••••••••

The first equation comes from the “Number of coins” column.

(Number of nickels) + (Number of quarters) = (total number of coins)

Equation: x + y = 28
—————————————————————
The second equation comes from the “value of all coins” column.

(Value of all nickels) + (Value of all quarters) = (Total value of all coins)

0.05x + 0.25y = 2.20

Remove the decimals by multiplying each term by 100:

5x + 25y = 220
—————————————————————
So we have the system of equations:

{x + y = 28
{5x + 25y = 202

Solve by substitution. Solve the first equation for y:

x + y = 28
y = 28 - x

Substitute (28 - x) for y in 5x + 25y = 220

5x + 25 (28 - x) = 220
5x + 700 - 25x = 220
-20x + 700 = 220
-20x = -480
x = 24

The number of nickels is 24.
————————————————————
Substitute in y = 28 - x
y = 28 - (24)
y = 4

The number of quarters is 4.
————————————————————
Checking:

24 nickels is $1.20 and 4 quarters is $1.00
That’s 28 coins.
Indeed $1.20 + $1.00 = $2.20
————————————————————

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Find the intervals on which​ f(x) is​ increasing, the intervals on which​ f(x) is​ decreasing, and the local extrema. f (x )equa
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f(x) is increasing in the intervals (-∞,-4) and (9,∞)

f(x) is decreasing in the ingercal (-4,9)

The local extrema is:

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and local min at (9,-1701)

Step-by-step explanation:

In order to solve this problem we must start by finding the derivative of the provided function, which we can find by using the power rule.

if f(x)=ax^{n} then f'(x)=anx^{n-1}

so we get:

f(x)=2x^{3}-15x^{2}-216x

f'(x)=6x^{2}-30x-216

in order to find the critical points we mus set the derivative equal to zero, since the local max an min will happen when the slope of the tangent line to the given point is zero, so we get:

6x^{2}-30x-216=0

we can solve this by factoring, so let's factor that equation:

6(x+4)(x-9)=0

we can now set each of the factors equal to zero so we get:

x+4=0 and x-9=0

when solving each for x we get that:

x=-4 and x=9

These are our critical points, now we can build the possible intervals we are going to use to determine where the function will be increasing and where it will be decreasing:

(-∞,-4), (-4,9) and (9,∞)

so now we need to test these intervals in the derivative to see if the graph will be increasing or decreasing in the given intervals. So let's pick x=-5 for the first one, x=0 for the second one and x=10 for the third one.

When evaluating them into the first derivative we get that:

f'(-5)=84, this is a positive answer so it means that the function is increasing in the interval (-∞,--4)

f'(0)=-216, this is a negative anser so it means that the function is decreasing in the interval (-4,9)

f'(10)=84, this is a positive answer so it means that the function is increasing in the interval (9,∞)

Now, for the local extrema, we can see that at x=-4, the function it's increasing on the left of this point while it's decreasing to the right, which means that there will be a local maximum at x=-4, so the local max is the point (-4,496)

We can see that at x=9, the function it's decreasing on the left of this point while it's increasing to the right, which means that there will be a local minimum at x=-9, so the local min is the point (9,-1701)

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