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marin [14]
2 years ago
9

A collection of 30 coins worth $5.50 consists of nickels, dimes and quarters. There are twice as many dimes as nickels. How many

quarters are there?
Mathematics
1 answer:
MAVERICK [17]2 years ago
4 0
<span>Let N equal the number of nickels.
Let D equal the number of dimes.
Let Q equal the number of quarters.

We know that .05N + .10D + .25Q = 5.50.
We also know that N + D + Q = 30.
We also know that D = 2N.

Therefore, Q = 30 - (N + D) = 30 - (N + 2N) = 30 - 3N

So...

5.50 = .05N + .10(2N) + .25(30 - 3N)
= .05N + .20N + 7.50 - .75N

Therefore (moving all N's to the left of the equals sing, and all constants to the right of the equals sign)...

.5N = 2.

Therefore, N = 4. There are four nickels, eight dimes, and 18 quarters.

Checking our work...
Four nickels is 0.20.
Eight dimes is 0.80.
18 quarters is 4.50.

That sums to 5.50, and is 30 coins.</span>
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8 is 16% of what?<br> Complete the sentence
zalisa [80]

Answer:

50

Step-by-step explanation:

the answer is 50 becuz it is

3 0
2 years ago
Christov and Mateus gave out candy to children on Halloween. They each have out candy at a constant rate, and they both gave awa
padilas [110]

Answer: 1) Both gave same number of candies to each child.

2) Christov gave candy to a greater number of children.

Step-by-step explanation:

Since, Christov initially had 300 pieces of candy, and after he was visited by 17 children, he had 249 pieces left.

Let he gave x candy to each student when he visited 17 children.

Then, 17 x + 249 = 300

⇒ 17 x = 51

⇒ x = 3

Since, he distributes candies in a constant speed.

Therefore, his speed of distribution = 3 candies per child

Now,The function that shows the remaining number of candies Mateus has after distributing candies to n children,

C(n)=270 - 3 n

Initially, n = 0

C(0) = 270

⇒ Mateus has initial number of candies = 270

When he gave candies to one child then remaining candies = 270 - 3 × 1 = 267

Thus, the candies, get by a child from Mateus = 270 - 267 = 3

Since, he distributes candies in a constant speed.

⇒ His speed of distribution = 3 candies per child

1) Therefore, Both Christov and Mateus have same speed of distribution.

2) Since, both have same seed.

⇒ The one who has greater number of candies will be distribute more.

⇒ Christov will give more candies.


8 0
3 years ago
Please help.<br> Is algebra.
Dvinal [7]

Answers:

  • Blank 1:  4
  • Blank 2:  -1

So the solution is (x,y) = (4, -1)

=============================================================

Work Shown:

6x + 7y = 17

6x + 7( y ) = 17

6x + 7( -3x+11 ) = 17 ... replace every copy of y with -3x+11

6x - 21x + 77 = 17

-15x = 17-77

-15x = -60

x = -60/(-15)

x = 4

We'll use this x value to find y

y = -3x+11

y = -3(4)+11 ... replace x with 4

y = -12+11

y = -1

We have  x = 4 and y = -1 pair up together to give us the solution (x,y) = (4, -1)

------------------------

To check the solution, we plug x = 4 and y = -1 into each equation

Plugging the values into the first equation leads to...

y = -3x+11

-1 = -3(4)+11

-1 = -1

This is effectively already done in the last part of the previous section. But it doesn't hurt to verify like this regardless.

We'll need to verify the second equation as well.

6x + 7y = 17

6(4) + 7(-1) = 17

24 - 7 = 17

17 = 17

We get a true equation, so the solution is confirmed with both equations. Overall, the solution to the system of equations is confirmed. This system is independent and consistent.

3 0
3 years ago
Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

\Rightarrow y(0.6)\approx 0.8800

Substituting x =0.8 and h= 0.2

y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

4 0
3 years ago
What is the rule for finding the coordinates of an image reflected over the line y=-x
joja [24]
You can find the coordinates of any graph by using a coordinate table, which looks as so in the chart provided below.

What you do is you, put in the letter x and then put what y is equal to in the second box. You plug in any number for x and you solve in order to find the y coordinate. Once you do, for example for plugging in -1 into x you get 1. So your coordinate would be (-1,1).

8 0
3 years ago
Read 2 more answers
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