Which expression can be used to calculate the rate per second at which the machine launches the balls?

What number can replace the n in n/12 to make it equivalent to 1/2 ?<br> 6<br> 8<br> 9<br> 10

Ierofanga [76]

Answer:

i believe the answer is 6.

4 0

3 years ago

Read 2 more answers

A sweet factory produces 5 different chocolate bars. the different flavors are always produced in the same proportions, for ever

IRISSAK [1]

a. There are 933 chocolate bars

b. There are 1200 chocolate bars

c. There are 1580 chocolate bars

<h3>How to calculate the number of chocolate bars</h3>

Since there are 5 different flavors,

let

x = orange flavored bars,
y = coconut flavored bars,
z = coffee flavored bars,
a = strawberry flavored bars and
b = honeycomb flavored bars and
X = total number of bars

Since the different flavors are always produced in the same proportions, for every 4 coconut flavored bars 5 honeycomb 6 orange 1 coffee and 4 strawberry

So, the ratio of their proportions are x:y:z:a:b = 6:4:2:1:5

So, the total ratio is T = 6 + 4 + 1 + 4 + 5 = 20

<h3>a. What is the total number of chocolate bars if there are 280 orange flavored bars?</h3>

Since we have 6 orange flavored bars, the ratio of orange flavored bars to total is 6/20

So, the amount of orange flavored bars is x = 6/20 × X

Making X subject of the formula, we have

X = 20x/6

So, if there are 280 orange bars, there will be

X = 20x/6

X = 20 × 280/6

X = 5600/6

X = 933.33

X ≅ 933 chocolate bars

So, there are 933 chocolate bars

<h3>b. What is the total number of chocolate bars if there are 960 coconut flavored bars?</h3>

Since we have 4 coconut flavored bars, the ratio of coconut flavored bars to total is 4/20

So, the amount of coconut flavored bars is y = 4/20 × X

Making X subject of the formula, we have

X = 20y/4

So, if there are 960 coconut flavored bars, there will be

X = 20y/6

X = 20 × 960/4

X = 5 × 240

X = 1200 chocolate bars

So, there are 1200 chocolate bars

<h3>c. What is the total number of chocolate bars if there are 79 coffee flavored bars?</h3>

Since we have 1 coffee flavored bars, the ratio of coffee flavored bars to total is 1/20

So, the amount of coconut flavored bars is z = 1/20 × X

Making X subject of the formula, we have

X = 20z

So, if there are 79 coffee flavored bars, there will be

X = 20z

X = 20 × 79

X = 1580 chocolate bars

So, there are 1580 chocolate bars

Learn more about ratio here:

brainly.com/question/1127546

#SPJ1

7 0

1 year ago

20% of the people in a village own a dog, 30% own goats and 40% own cattle, 5% own both a dog and goats, 4% own both a dog and c

lana66690 [7]

Answer:

37 %

Step-by-step explanation:

here..

only Dog = 20- 5 - 4 - 1 = 10

only goats = 30 - 5 - 3 - 1 = 21

only cattle = 40 - 4 - 3 - 1 = 32

then ... since the maximum percentage is 100 %

people who don't own any animal = 100% - 10% - 21% - 32% = 37 %

3 0

3 years ago

I'm stuck on 14-16 please help

sukhopar [10]

14 - 16 is -2.

Hope this helped.

3 0

3 years ago

Use lagrange multipliers to find the point on the plane x â 2y + 3z = 6 that is closest to the point (0, 2, 4).

Arisa [49]

The distance between a point $(x,y,z)$ on the given plane and the point (0, 2, 4) is

$\sqrt{f(x,y,z)}=\sqrt{x^2+(y-2)^2+(z-4)^2}$

but since $\sqrt{f(x,y,z)}$ and $f(x,y,z)$ share critical points, we can instead consider the problem of optimizing $f(x,y,z)$ subject to $x-2y+3z=6$ .

The Lagrangian is

$L(x,y,z,\lambda)=x^2+(y-2)^2+(z-4)^2+\lambda(x-2y+3z-6)$

with partial derivatives (set equal to 0)

$L_x=2x+\lambda=0\implies x=-\dfrac\lambda2$
$L_y=2(y-2)-2\lambda=0\implies y=2+\lambda$
$L_z=2(z-4)+3\lambda=0\implies z=4-\dfrac{3\lambda}2$
$L_\lambda=x-2y+3z-6=0\implies x-2y+3z=6$

Solve for $\lambda$ :

$x-2y+3z=-\dfrac\lambda2-2(2+\lambda)+3\left(4-\dfrac{3\lambda}2\right)=6$
$\implies2=7\lambda\implies\lambda=\dfrac27$

which gives the critical point

$x=-\dfrac17,y=\dfrac{16}7,z=\dfrac{25}7$

We can confirm that this is a minimum by checking the Hessian matrix of $f(x,y,z)$ :

$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

$\mathbf H$ is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.

At this point, we get a distance from (0, 2, 4) of

$\sqrt{f\left(-\dfrac17,\dfrac{16}7,\dfrac{25}7\right)}=\sqrt{\dfrac27}$

8 0

3 years ago