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joja [24]
3 years ago
8

375 of the students at Haldane Elementary School walk to school. If these

Mathematics
1 answer:
ELEN [110]3 years ago
7 0

Answer:

506

Step-by-step explanation:

do the math

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Solve for the value of x: x/2 + x/3 = 2x/5 + 3 , also I need a very thorough explanation for this.
Alona [7]

Answer:

x=\frac{90}{13} .

Step-by-step explanation:

\frac{x}{2} +\frac{x}{3} =\frac{2x}{5} +3;

30*(\frac{x}{2} +\frac{x}{3} )=30*(\frac{2x}{5} +3);

15x+10x=12x+90; ⇔ 13x=90;

x=\frac{90}{13}.

8 0
2 years ago
Write the equation of each line in slope-intercept form and identify the slope. 4x=2+y
RideAnS [48]

Answer:

y = 4x - 2

Step-by-step explanation:

Given: 4x=2+y

First, subtract 2 on both sides to isolate out y. (Slope-intercept form is y=mx+b)

Once subtracted, you get: 4x-2=y

Rearrange it, and you get: y=4x-2

7 0
2 years ago
Does the graph represent a function?
Musya8 [376]

Answer:

yes

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
What is the simplified form of 1/x-1/y/1/x+1/y
Svetradugi [14.3K]
Is this the expression to simplify?

8 0
3 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
7 0
3 years ago
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