All categories

2 years ago

Noah finds some nickels and pennies in his change purse. How many coins does he

Mathematics

1 answer:

zzz [600]2 years ago

8 0

He has 100 + 120 = 220 coins in his purse.

He has n + p coins if he has n nickels and p pennies.

You might be interested in

What is 1/2(7x-6)=6x-10

erastova [34]

Answer:

Step-by-step explanation:

1/2(7x - 6) = 6x - 10

7/2x - 3 = 6x - 10 .....multiply by 2

7x - 6 = 12x - 20

-6 + 20 = 12x - 7x

14 = 5x

14/5 = x <===

7 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! Graph g(x) = 3x^2 - 12x - 3

Paul [167]

Answer: Vertex = (2, -15) 2nd point = (0, -3)

Step-by-step explanation:

g(x) = 3x² - 12x - 3

= 3(x² - 4x - 1)

a=1 b=-4 c=-1

Find the x-value of the vertex by using the formula for the axis of symmetry: $x = \dfrac{-b}{2a}$

$x = \dfrac{-(-4)}{2(1)}$

$= \dfrac{4}{2}$

= 2

Find the y-value of the vertex by plugging the x-value (above) into the given equation: g(x) = 3x² - 12x - 3

g(2) = 3(2)² - 12(2) - 3

= 12 - 24 - 3

= -15

So, the vertex is (2, -15) ← PLOT THIS COORDINATE

Now, choose a different x-value. Plug it into the equation and solve for y. I chose x = 0

g(0) = 3(0)² - 12(0) - 3

= 0 - 0 - 3

= -3

So, an additional point is (0, -3) ← PLOT THIS COORDINATE

5 0

3 years ago

To rent a bounce house: At Mike's Bounce Shop costs $30 plus $2 per hour. At Jose's Bounce Rentals, it costs $12 plus $8 per hou

ANEK [815]

Answer: After 3 hours both will have the same cost.

Step-by-step explanation:

Let x = Number of hours and y be the total cost.

Total cost = Initial cost + (cost per hour)(Number of hours)

At Mike's Bounce Shop,

Total cost (y)= 30+2x

At Jose's Bounce Rentals,

Total cost(y) = 12+8x

When both shops have the same cost, then

$12+8x=30+2x\\\\\Rightarrow\ 8x-2x=30-12\\\\\Rightarrow\ 6x=18\\\\\Rightarrow\ x=3$

Total cost = 12+8(3)=12+24=$36

Hence, After 3 hours both will have the same cost.

3 0

3 years ago

Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices

Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

$\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt$

Where P, Q are scalar functions

We want to compute

$\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy$

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths $\large C_1,C_2,C_3,C_4$ which represents the sides of the rectangle.

$\large C_1$ is the line segment from (0,0) to (5,0)

$\large C_2$ is the line segment from (5,0) to (5,1)

$\large C_3$ is the line segment from (5,1) to (0,1)

$\large C_4$ is the line segment from (0,1) to (0,0)

Then

$\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}$

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize $\large C_1$ as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

$\large \displaystyle\int_{C_1}xydx+x^2dy=0$

Now the second integral. We parametrize $\large C_2$ as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

$\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25$

The third integral. We parametrize $\large C_3$ as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

$\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2$

The fourth integral. We parametrize $\large C_4$ as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

$\large \displaystyle\int_{C_4}xydx+x^2dy=0$

$\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2$

Now, let us compute the value using Green's theorem.

According with this theorem

$\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx$

where A is the interior of the rectangle.

so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}

We have

$\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x$

$\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2$

3 0

3 years ago

I accidentally click on the wrong subscriptions like I want the 1 month but I got the 12 months. Is there anyway I could contact

Yanka [14]

Answer:

Step-by-step explanation:

Scroll down to the bottom of the screen and click on contact us

4 0

3 years ago