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3 years ago

What is −1/2 − (−4/5) = x Please Help Its due in like 15 minutes

Mathematics

2 answers:

laila [671]3 years ago

8 0

Answer:

x = 3/10

Step-by-step explanation:

−1/2 − (−4/5) = x

First, switch sides.

x = −1/2 − (−4/5)

You are subtracting a negative number. Subtracting a negative number means add a positive number.

x = -1/2 + 4/5

Now you are adding two fractions, so you need a common denominator. The LCD of 2 and 5 is 10.

x = -5/10 + 8/10

x = 3/10

dimulka [17.4K]3 years ago

8 0

Answer:

your answer might be 1.1

Step-by-step explanation:

You see you have to divide it so it will be -1 divided by 2-(- 4 divided by5)=x so if we do x - those all our answer will be 1.1 your welcome I hope that helps

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A shop sells chocolate, vanilla and strawberry ice creams. 45% of the ice creams sold are vanilla, and 1/4 are strawberry. What

Alinara [238K]

Answer:

Step-by-step explanation:

Total percentage = 100%

And we know that proportions of Vanilla +chocolate +strawberry = 100%

Proportion of Vanilla = 45%

Proportion of Strawberry = 1/4 *100 = 25%

Let proportion of chocolate be represented by variable C;

So, 45% +25% + C = 100%

65% +C = 100%

Subtract 65% from both sides to solve for C;

C=100% - 65%

C = 35%

Therefore there is 35% of chocolate in the shop.

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2 years ago

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3 years ago

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

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3 years ago

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san4es73 [151]

Answer:

Option D y=-3x^2 +4

Step-by-step explanation:

This is the correct answer because the graph is going down which shows that it is negative and the y intersect is +4

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