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belka [17]
2 years ago
7

Help with this question ??

Mathematics
1 answer:
NNADVOKAT [17]2 years ago
5 0

Answer: 12 tables minimum, 15 max.

Step-by-step explanation:

If you substitute 14 in an inequality

200c + 500t >= 8800

200(14) + 500t and solve for t, you get t must be at least 12.

C +T cannot exceed 29

So figure 14 +12 =26 so they could sell up to 15 tables and not go over 29.

You will have to enter the t values 12,13,14,15

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Is sqrt(25a^2/4) the same as sqrt(25a^2)/sqrt(4)?
Evgesh-ka [11]
LHS\\ \\ =\sqrt { \frac { 25{ a }^{ 2 } }{ 4 }  } \\ \\ ={ \left( \frac { 25{ a }^{ 2 } }{ 4 }  \right)  }^{ \frac { 1 }{ 2 }  }

\\ \\ =\frac { { \left( 25{ a }^{ 2 } \right)  }^{ \frac { 1 }{ 2 }  } }{ { 4 }^{ \frac { 1 }{ 2 }  } } \\ \\ =\frac { \sqrt { 25{ a }^{ 2 } }  }{ \sqrt { 4 }  } \\ \\ =RHS
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3 years ago
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A quadrilateral is dilated by a scale factor of 2. It's original side measures are 7, 14, 21, and 28.
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Answer:

14,28,42 and 56

Step-by-step explanation:

Multiply all lengths by 2 as the quadrilateral is dilated by a scale factor of 2

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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?
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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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Round 768 to the nearest 10
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SOLVE: High school precal
allochka39001 [22]
Tbh idk I just need help passing a class
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