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James has a bag 77 of nickels and dimes. The total value of coins is $6.60. How many dimes does James have?

tatuchka [14]

James has 55 dimes in the bag

Step-by-step explanation:

The given is:

James has in a bag 77 of nickels and dimes
The total value of coins is $6.60

We need to find how many dimes James has

Assume that the number of nickels is x and the number of

dimes is y in the bag

∵ There are x nickels in the bag

∵ There are y dimes in the bag

∵ There are 77 coins in the bag

∴ x + y = 77 ⇒ (1)

∵ 1 nickel = 5 cents

∴ The value of nickles in the bag = 5x cents

∵ 1 dime = 10 cents

∴ The value of dimes in the bag = 10y cents

∵ The total value of coins is $6.60

∵ 1 dollar = 100 cents

∴ $6.60 = 6.60 × 100 = 660 cents

∴ 5x + 10y = 660 ⇒ (2)

Now we have a system of equations to solve it

Multiply equation (1) by -10 to eliminate y

∵ -10x - 10y = -770 ⇒ (3)

- Add equations (2) and (3)

∴ -5x = -110

- Divide both sides by -5

∴ x = 22

Substitute the value of x in equation (2) to find y

∵ 22 + y = 77

- Subtract 22 from both sides

∴ y = 55

∴ The number of dimes is 55

James has 55 dimes in the bag

Learn more:

You can learn more about the system of equations in brainly.com/question/6075514

#LearnwithBrainly

6 0

4 years ago

A bus driver pays all tolls, using only nickels and dimes, by throwing one coin at a time into the mechanical toll collector.

lutik1710 [3]

Answer: a) An = An-1 + An-2

b) 55ways

Step-by-step explanation:

a) a nickel is 5 cents and a dime is 10cent so a multiple of 5 cents is the possible way to pay the tolls in both choices.

Let An represents the number of possible ways the driver can pay a toll of 5n cents, so that

An = 5n cents

Case 1: Using a nickel for payment which is 5 cents, the number of ways given as;

An-1 = 5( n-1)

Case 2: using a dime which is two 5 cents, the number of ways is given as;

An-2 = 5(n-2)

Summing up the number of ways, we have

An = An-1 + An-2

From the relation,

If n= 0, Ao= 1

n =1, A1= 1

b) 45 cents paid in multiples of 5cents will give us 9 ways(A9)

From the relation, we have that

Ao = 1

A1 = 1

An =An-1 + An-2

Ao = 1

A1 = 1

A2 = A1+Ao = 1+1= 2

A3 = A2 + A1 = 3

A4 = A3+A2=5

A5=A4+A3=8

A6=A5+A4=13

A7 =A6+A5 = 21

A8= A7+A6= 34

A9= A8+A7= 55

So there are 55ways to pay 45cents.

4 0

3 years ago

ABCD is a parallelogram. If mZCDA = 63, then mZABC = _?_ The diagram is not to scale.

nevsk [136]

Answer:

b. 63

Step-by-step explanation:

A Parallelogram is known to be a flat shape quadrilateral which has four opposite sides Each two opposite sides are parallel and are equal in length. The opposite angles are equal.

The above answer is correct because opposite angles of a parallelogram are equal:

m∠CDA = m∠ABC

where m∠CDA = 63

∴ m∠ABC = 63 (answer)

4 0

3 years ago

50 POINTS, PLEASE ANSWER

ludmilkaskok [199]

Step-by-step explanation:

Part 1:

Rotation counterclockwise 90 &degree; of a point is

Here is a table of the transformation of all four points:

P(-7,0)-> P'(-0,-7) =P'(0,-7)

A(-4,2)-> A'(-2,-4)

R(-3,0)-> R'(0,-3)

K(-6,2)-> K'(-2,-6)

The initial and rotated images are shown attached.

Part 2: it needs to be done with the technology tool that you have. We can check your work if you post what you did.

7 0

3 years ago

Read 2 more answers

a line passes through (3,-2) and (6,2). write an equation in point-slope form. rewrite the equation in standard form

Vlad [161]

again, bearing in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{2}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-(-2)}{6-3}\implies \cfrac{2+2}{6-3}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-2)=\cfrac{4}{3}(x-3)\implies y+2=\cfrac{4}{3}x-4 \\\\[-0.35em] ~\dotfill$

$\bf y=\cfrac{4}{3}x-6\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y)=3\left( \cfrac{4}{3}x-6 \right)}\implies 3y=4x-18 \\\\\\ -4x+3y=-18\implies \stackrel{\textit{standard form}}{4x-3y=18}$

6 0

4 years ago