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Maksim231197 [3]
3 years ago
14

What is the area of triangle pqr on the grid? a triangle pqr is shown on a grid. the vertex p is on ordered pair 7 and 6, vertex

q is on ordered pair 1 and 6, and the vertex r is on ordered pair 4 and 2. 5 square units 6 square units 10 square units 12 square units?
Mathematics
2 answers:
aivan3 [116]3 years ago
7 0

Answer:

12

Step-by-step explanation:

horsena [70]3 years ago
6 0

P(7,6), Q(1,6), R(4,2)

We have PQ parallel to the x axis. We'll call that the base,

b = 7 - 1 = 6

The altitude is then the y difference h = 6 - 2 = 4

The area is \frac 1 2 bh = \frac 1 2 (6)(4) = 12

Answer: 12 square units


In general we can use the shoelace formula for the area of any polygon given coordinates. We write the points like this:

(7,6), (1,6), (4,2)

(1,6), (4,2), (7,6)

The area is then half the absolute value of the sum of the cross products:

A = \frac 1 2 | 7(6)-6(1) + 1(2)-6(4) + 4(6)-2(7) | = \frac 1 2 |24| = 12


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A state lottery game consists of choosing one card from each of the four suits in a standard deck of playing cards. (There are 1
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Answer:

17160

Step-by-step explanation:

first there are 13 options

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then there are 13 - 2 that were chosen

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little further explaining of why this works:

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the answer is 6 which is also 3 * 2 * 1 = 6

HOPE THAT HELPS!1! ^_^

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3 years ago
What are the coordinates of the vertex for f(x) = 3x2 + 6x + 3?
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A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int
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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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All except for h(x) are exponential growth because it is over 1
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