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Paraphin [41]
3 years ago
6

You bought a magazine for $5 and four erasers. You spent a total of $25. How much did each eraser cost?

Mathematics
1 answer:
Novay_Z [31]3 years ago
7 0
Well all you would have to do is subtract 25 to 5 which is 5 so $20 is how much
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Increase $32 in the ratio 2:5<br>decrease 54 litter in the ratio 5:1<br>​
Novay_Z [31]

Answer:

Step-by-step explanation:

32 = 2 : 5

32/7 = 4

2 x 4 : 4 x 5

8 : 20

8 0
2 years ago
Gary made 5 baskets than Susan this season bill made 3 less then twice as many baskets as Susan. Together, all three made 42 bas
Nezavi [6.7K]

Answer:Gary made 15 baskets.

Susan made 10 baskets.

Bill made 17 baskets.

Step-by-step explanation:

Let x represent the number of baskets that Gary made.

Let y represent the number of baskets that Susan made.

Let z represent the number of baskets that Bill made.

Gary made 5 baskets than Susan this season. This means that

x = y + 5

Bill made 3 less than twice as many baskets as Susan. This means that

z = 2y - 3

Together, all three made 42 baskets. This means that

x + y + z = 42 - - - - - - - - - - 1

Substituting x = y + 5 and z = 2y - 3 into equation 1, it becomes

y + 5 + y + 2y - 3 = 42

4y + 2 = 42

4y = 42 - 2 = 40

y = 40/4 = 10

x = y + 5 = 10 + 5

x = 15

z = 2y - 3 = 2 × 10 - 3

z = 20 - 3 = 17

3 0
3 years ago
Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g
AysviL [449]

Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

The initial concentration in the tank is:

c_{o} = \frac{10\,pd}{60\,gal}

c_{o} = 0.167\,\frac{pd}{gal}

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

7 0
3 years ago
I’m stuck with this question :(
gayaneshka [121]

Answer:

The same number was not added to both sides.

Step-by-step explanation:

Given:

t - 8 = 65

in the next step it states something wrong, because it should have been this:

t - 8 = 65

add +8 left and right of the = sign.

t - 8 + 8 = 65 + 8

t + 0 = 73

t = 73

So the error made what's that

The same number was not added to both sides.

8 0
3 years ago
Read 2 more answers
Please help i will give brainliest
inysia [295]

Answer:

Step-by-step explanation:

Take the area of the yard and subtract from it the area of the pool.  In quadratic form, the area of the pool is

2x^2-x-28

Subtracting the area of the pool from the area of the yard:

3x^2-14x-5-(2x^2-x-28)

Since the negative in front of the parenthesis will change the signs inside:

3x^2-14x-5-2x^2+x+28

Combine like terms to get the area left after the pool goes in:

x^2-13x+23

8 0
3 years ago
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