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umka21 [38]
3 years ago
5

I need help with this

Mathematics
1 answer:
Wewaii [24]3 years ago
3 0

Answer:

12x+24x+18 and 36x+24-6 can both be simplified into 36x+18, meaning the equation is an identity

You might be interested in
What are the zeros of the quadratic function f(x) = 2x2 + 8x – 3?
julsineya [31]
The zeros of the function are also called the roots. Thus, this problem is basically locating the root of the given equation.

To locate the roots, we equate the equation f(x) to 0.

We then dissect each term to identify a, b, and c in the quadratic equation formula:

x = (- b +- √(b²-4ac))/2a

a = 2
b = 8
c = -3

x1 = 0.3452
x2 = -4.345


8 0
3 years ago
Read 2 more answers
5. Using the information from problem 4, if the club sells 40 tickets in advance, what is the least number the club needs to sel
larisa [96]

Answer:

8

Step-by-step explanation:

40 + 5 (at the door)=45

(president's goal) 400 divided by 45= approximately 8

4 0
3 years ago
PLEASE ANSWER QUICKLY! WILL GIVE BRANLIEST
uysha [10]

Answer:

Step-by-step explanation:

Let x represent the seating capacity

Number of seats = 40+x

Profit per seat = 10 - 0.20x

For maximum number of seats

P(x) = ( 40+x ) ( 10-0.20x )

P(x) = 400+10x-8x-0.2x^2

P(x) = 400+2x- 0.2x^2

Differentiating with respect to ( x )

= 2 - 0.4x

0.4x = 2

x = 2/0.4

x = 5

The seating capacity will be 40+5 = 45

For the maximum profits

40X10+ 9.9 + 9.8 + 9.7 + 9.6 + 9.5 + 9.4 + 9.3 + 9.2 + 9.1 + ... 1.0, 0.9, 0.8, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1

= 400 + an arithmetic series (first term = 0.1, common difference = 0.1, number of terms = 8+ 40 = 48 )

= 400 + (48/2)(2X0.1 + (48-1)X0.1)

= 400 + 24(0.2 + 4.7)

= 400 + 24(4.9)

= 400 + 117.6

= 517.6

= 517.6dollars

3 0
3 years ago
10a^2b^0 for a=–3, b=–8
NeTakaya

Answer:

<h2>90</h2>

Step-by-step explanation:

Substitute a = -3 and b = -8 to the expression 10a²b⁰:

10(-3)²(-8)⁰ = 10(9)(1) = 90

a⁰ = 1 for any real number

7 0
3 years ago
Read 2 more answers
A geometric sequence has first term 1/9 and common ratio 3. Which is the first term of the sequence which exceeds 1000?
Svetach [21]
a_{n}= \frac{1}{9}  (3)^{n-1}
(We know this from a=1/9 and r=3)
Simplifying this, we get:
\frac{1}{9} (3)^{-1} (3)^n

Since we're finding the first term that exceeds 1000, let's set it equal to 1000.

\frac{1}{27}(3)^n=1000
Multiplying both sides by 27
3^n=27000

log_{3}27000=n

n≈9.2

We have to round n up, since if n=9, the value would be <1000.
Therefore n=10. Substituting n=10,
\frac{1}{27}3^{10}
=2187

Therefore the first term that exceeds 1000 is 2187, and it is the 10th term
3 0
3 years ago
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