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polet [3.4K]
2 years ago
8

VIE An expression is shown. 5-3+1 -4 What is the value of the expression?

Mathematics
1 answer:
Citrus2011 [14]2 years ago
4 0
The value is negative one (-1)
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Find the least common denominator for these two rational expressions. B/b^2 -64 -7b/b^2+7b-8
san4es73 [151]

Answer:

The least common denominator is (b-8)(b+8)(b-1)

Step-by-step explanation:

We are given expression as

\frac{b}{b^2-64}-\frac{7b}{b^2+7b-8}

Firstly, we will factor both denominators

b^2-64=b^2-8^2=(b-8)(b+8)

b^2+7b-8=(b+8)(b-1)

so, we can plug it back

\frac{b}{(b-8)(b+8)}-\frac{7b}{(b+8)(b-1)}

First term denominator is

(b-8)(b+8)

Second term denominator is

(b+8)(b-1)

So,

Least common denominator will be

(b-8)(b+8)(b-1)

So, we get

LCD=(x-8)(x+8)(x-1)


3 0
2 years ago
Everbank Field, home of the Jacksonville Jaguars, is capable of seating 76,867 fans. The revenue for a particular game can be mo
k0ka [10]
The total revenue for the event would the total amount earned from the tickets sold. So if x people attended the event, then there would be $(161x). We must keep in mind though the the maximum seating is 76 876. That means that the maximum revenue that can be earned must not exceed $(161)(76 867) = . Hence we have f(x) = 161x where 0 ≤ x ≤ 76867. From this, we can see that the domain is [0, 76867] while range is [0, 12375587]<span>.</span>
6 0
3 years ago
Please help me <br> 3(x + 2) * 10 * 7 divided by 70 - 2 (x+2)<br><br> Simplify
I am Lyosha [343]

Answer:

= 105(x+2)/33-x

Step-by-step explanation:

Given the expression

[3(x+2)*10*7]÷70 - 2(x+2)

= 3(x+2)*70÷70 - 2(x+2)

= 210(x+2)/70-2x-4

= 210(x+2)/66-2x

= 210(x+2)/2(33-x)

= 105(x+2)/33-x

8 0
2 years ago
What time is it 7 in a half hours before 2:12 am
scoray [572]
6.42 pm is the answer. You can think backward 8 hours before 2:12 am, which is 6:12pm. Now, go 'forward' a half an hour<span>, to 6:42pm. Answer: 6:42pm.</span>
5 0
3 years ago
I need help on this one guys :(
Sophie [7]

Answer:

the last one

Step-by-step explanation:

6 0
2 years ago
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