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Andrei [34K]
2 years ago
7

Which equation does the graph of the systems of equations solve? 2 linear graphs. They intersect at 1,4

Mathematics
1 answer:
Pepsi [2]2 years ago
6 0

Answer:

See below.

Step-by-step explanation:

There is an infinite n umber of systems of equations that has (1, 4) as its solution. Are you given choices? Try x = 1 and y = 4 in each equation of the choices. The set of two equations that are true when those values of x and y are used is the answer.

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A cone has a volume of 942 in.³ and a height of 9 inches what is the radius of the cone use 3.14 for pie
bija089 [108]

Answer:

r = 51.96 inches

Step-by-step explanation:

V_{cone} = \pi {r}^{2} h  \\  \\  \implies \: r^{2}  =  \frac{V_{cone}}{\pi h } \\  \\  \implies \:  r^{2} =  \frac{942}{3.14 {(9) } }\\  \\  \implies \:  r^{2} =  \frac{942}{3.14 {(9) } } \\  \\  \implies \:  r^{2} =2700 \\  \\  \implies \:  r= \sqrt{2700}  \\  \\  \implies \:  r=51.96 \: inches

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2 years ago
For the given equation, find the values of a, b, and c, determine the direction in which
Shkiper50 [21]

Answer: Table D

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  • This eliminates tables A, B, and C.

Thus, table D is the answer.

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2 years ago
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
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