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Ivenika [448]
2 years ago
9

Algebra Vocabulary Crossword help ?!

Mathematics
2 answers:
skad [1K]2 years ago
6 0
1.) Function Notation
9.) Linear Function
avanturin [10]2 years ago
5 0
1. a relation described in word format (17 letter) is <span> "is direct variation"

9.  a way to name a function identifying its input (16 letter) is "</span><span>f of x equals x square"
</span>
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A racquetball club charges $10 for a one-month trial membership. After the trail month, the regular membership fee is $12 per mo
Furkat [3]
Membership is discounted $2 for first month
c=cost of one month
d=discount=$2
x=# of months
xc - 2= total cost
x for 1 month is 1, c is always 12, -2 is constant
1 x(12) - 2=10 for first month
8 months
8c -2=total cost for 8 months
8 x 12 - 2= 94 for 8 months

5 0
3 years ago
Nicole bought six pounds of apples at $1.50 per pound. The store had a discount of $2 off her total purchase. Nicole and her fri
Andru [333]

Answer:

Step-by-step explanation:

Total amount of apples that Nicole bought was 6 pounds.

Each pound of apple cost $1.50

This means that the total cost of the 6 pounds of apple that Nicole bought would be

6 × 1.5 = $9

The store had a discount of $2 off her total purchase. This means that the amount that was paid is

9 - 2 = $7

If Nicole and her friend then divided the cost of the purchase evenly, let x represent the amount that each of them would pay. Therefore, the expression that can be used to determine how much Nicole and her friend each paid for the apples would be

x = 7/2

6 0
3 years ago
Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
find the average of the numbers 5/8, 9/10, 2 3/4 write the solution as a mixed number or fraction in lowest terms.​
oksano4ka [1.4K]

Answer:

3.025, 121/40

Step-by-step explanation:

3 0
2 years ago
Find (f o f)(0). f(x)=x^2-1
Anuta_ua [19.1K]

Answer:

The function operation is 0

The composite function is -1

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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