Please help me I need help

8. myles had 103 coins in a jar - all nickels, dimes, and quarters. if the total value of the coins is $15

Ivanshal [37]

If the total value of the coins is $15. The number of each type of coin is: 31 quarters, 72 dimes.

<h3>Number of each type of coin</h3>

Let D = number of dimes

Let Q = the number of quarters

Equations

d + q = 103

0.10d + 0.25q = 15

d = 103 -q

0.10(103 -q) + 0.25q = 15

10.3 - 0.10q + 0.25q = 15

0.15q = 4.7

q=4.7/0.15

q=31 quarters

Substitute q into second equation

D+31=103

D=72 dimes

Therefore the number of each type of coin is: 31 quarters, 72 dimes.

Learn more number of each coin here:brainly.com/question/13934075

#SPJ1

5 0

2 years ago

Please help ASAP!!! Step by step <br> translation : 3units right , 4 units down

xenn [34]

Answer:

C'=(-1,0)

B'=(-2,1)

A'=(-4,1)

D'=(-4,0)

Step-by-step explanation:

each point moves 3 right and 4 down ( follow the pattern)

Hope i helped :)

6 0

3 years ago

If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

2

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

3 years ago

Find the commission on a sale of $8,432 at 1 1/2% commission.

IrinaK [193]

Answer:

5621.33333333

Step-by-step explanation:

This is only the right answer if you are asking how much money 1 1/2 % of 8432 is.

But i think i'm right ^-^

6 0

3 years ago

The path of a roller coaster over a particular length of track can be represented by a quadratic function f, where f(x) represen

alexgriva [62]

The roller coaster could be represented by the function for Choice D.

If you were to graph this function, you would find that the zeros (x-intercepts) of the function are at the points of -2 and 1. Those are the needed points in the problem.

Since this is a parabola that opens upward, the parabola must be below the axis (or support bar) between those values.

8 0

3 years ago