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inysia [295]
3 years ago
7

Running at the same constant rate, 5 identical machines can produce one lot containing a certain number of toys per hour. At thi

s rate, how many toys could 8 machines produce in 1/2 hours?
Mathematics
1 answer:
inessss [21]3 years ago
6 0

Answer:

1.6\cdot{x}

Step-by-step explanation:

If the speed is constant for 5 machines and produces x amount of toys we know that rate is defined as a constant per time:

Here the machines are constant and the rate is x toys per hour:

If 5 machines make x per hour:

Rate is x/1 hour

The the ratio of 8/5 is 1.6. Therefore for 8 machines the rate is 1.6 times more per hour than for5 machines. Therefore we can write an expression as:

Rate for 5 machines is x and for 8 machines is y:

y=1.6\cdot{x}

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TiliK225 [7]
There are 3 parallel faces shown
5 0
3 years ago
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Al trazar en el plano cartesiano el angulo a cuyo lado terminal es el punto de ( 3,4). Los valores de las funciones del coseno y
Igoryamba

Answer:

cos(\alpha)=\frac{3}{5}=0.6

cosec(\alpha)=\frac{5}{4}=1.25

Step-by-step explanation:

El cos(α) se define como el cociente entre el cateto adyacente y la hipotenusa.

El valor del cateto adyacente en nuestgro caso es CA = 3.

La hipotenusa se calcual de la siguiente manera:

h=\sqrt{3^2+4^2}=5

Por lo tanto, el cos(α) sera:

cos(\alpha)=\frac{3}{5}=0.6

El cosec(α)=h/CO.

El cateto opuesto CO = 4 y la hipotenusa h = 5

Por lo tanto, el cosec(α) sera:

cosec(\alpha)=\frac{5}{4}=1.25

Espero te haya sido de ayuda!

4 0
3 years ago
Austin is on the swim team. Each week he swims a total of 3000 meters. How many kilometers does he swim each week?
marishachu [46]

Answer:

3 km

Step-by-step explanation:

<em>(NOTE: One kilometer is 1000 meters)</em>

Divide 3000 by 1000 to get 3

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=4x%20%2B%203y%20%3D%200%20%5C%5C5y%20%2B%2053%20%3D%2011x%20%5C%5C%20" id="TexFormula1" title=
Anvisha [2.4K]

Answer:

x = 3, y = -4

Step-by-step explanation by substitution:

Solve the following system:

{4 x + 3 y = 0 | (equation 1)

5 y + 53 = 11 x | (equation 2)

Express the system in standard form:

{4 x + 3 y = 0 | (equation 1)

-(11 x) + 5 y = -53 | (equation 2)

Swap equation 1 with equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

4 x + 3 y = 0 | (equation 2)

Add 4/11 × (equation 1) to equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+(53 y)/11 = -212/11 | (equation 2)

Multiply equation 2 by 11/53:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+y = -4 | (equation 2)

Subtract 5 × (equation 2) from equation 1:

{-(11 x)+0 y = -33 | (equation 1)

0 x+y = -4 | (equation 2)

Divide equation 1 by -11:

{x+0 y = 3 | (equation 1)

0 x+y = -4 | (equation 2)

Collect results:

Answer: {x = 3 , y = -4

_____________________________________

Solve the following system:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for x:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Isolate terms with x to the left hand side.

Subtract 3 y from both sides:

{4 x = -3 y

5 y + 53 = 11 x

Hint: | Solve for x.

Divide both sides by 4:

{x = -(3 y)/4

5 y + 53 = 11 x

Hint: | Perform a substitution.

Substitute x = -(3 y)/4 into the second equation:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for y:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Isolate y to the left hand side.

Subtract 53 - (33 y)/4 from both sides:

{x = -(3 y)/4

(53 y)/4 = -53

Hint: | Solve for y.

Multiply both sides by 4/53:

{x = -(3 y)/4

y = -4

Hint: | Perform a back substitution.

Substitute y = -4 into the first equation:

Answer: {x = 3 , y = -4

7 0
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A because 2b is b times 2 and 235 is bigger
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