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Nat2105 [25]
2 years ago
5

1. What values of a, b, and c would you use in the quadratic formula for the following equation?

Mathematics
1 answer:
ivolga24 [154]2 years ago
7 0

Answer:

D

Step-by-step explanation:

because it is the answer that I think of

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What is the product of − 3/11 and − 2/15? please explain.
emmainna [20.7K]

ANSWER

The product is

\frac{2}{55}

EXPLANATION

To find the product of

-  \frac{3}{11}

and

\frac{ - 2}{15}

means we should multiply the two fractions.

We multiply to obtain,

-  \frac{3}{11}  \times  -  \frac{2}{15}

=  -  \frac{3}{11}  \times  -  \frac{2}{3 \times 5}

Cancel the common factors:

=  -  \frac{1}{11}  \times  -  \frac{2}{5}

Now, multiply the numerators separately and the denominators too separately.

=  \frac{ - 1 \times  - 2}{11 \times 5}

This simplifies to,

= \frac{2}{55}

4 0
2 years ago
What is nine tenths in standard form?<br> I know it's simple but I forgot.
Fiesta28 [93]
That would be 0.9 because 9 is in the tenths place
5 0
3 years ago
How many solutions does the equation −5a + 5a + 9 = 8 have?
faust18 [17]

Answer: the answer will be none

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Tom wants to get to the playground at 2:30 p.m. it takes him 27 minutes to bike there
N76 [4]

Answer:

2:03

Step-by-step explanation:

If it takes him 27 minutes he should leave the house 27 minutes before he has to go. 2:03 is 27 minutes early

6 0
2 years ago
Read 2 more answers
A company manufactures and sells x television sets per month. The monthly cost and​ price-demand equations are ​C(x)=74,000+80x
SOVA2 [1]

Answer:

a) $675000

b) $289000 profit,3300 set, $190 per set

c) 3225 set, $272687.5 profit, $192.5 per set

Step-by-step explanation:

a) Revenue R(x) = xp(x) = x(300 - x/30) = 300x - x²/30

The maximum revenue is at R'(x) =0

R'(x) = 300 - 2x/30 = 300 - x/15

But we need to compute R'(x) = 0:

300 - x/15 = 0

x/15 = 300

x = 4500

Also the second derivative of R(x) is given as:

R"(x) = -1/15 < 0 This means that the maximum revenue is at x = 4500. Hence:

R(4500) = 300 (4500) - (4500)²/30 = $675000  

B) Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 80x) = -x²/30 + 300x - 80x - 74000

P(x) = -x²/30 + 220x - 74000

The maximum revenue is at P'(x) =0

P'(x) = - 2x/30 + 220= -x/15 + 220

But we need to compute P'(x) = 0:

-x/15 + 220 = 0

x/15 = 220

x = 3300

Also the second derivative of P(x) is given as:

P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3300. Hence:

P(3300) =  -(3300)²/30 + 220(3300) - 74000 = $289000  

The price for each set is:

p(3300) = 300 -3300/30 = $190 per set

c) The new cost is:

C(x) = 74000 + 80x + 5x = 74000 + 85x

Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 85x) = -x²/30 + 300x - 85x - 74000

P(x) = -x²/30 + 215x - 74000

The maximum revenue is at P'(x) =0

P'(x) = - 2x/30 + 215= -x/15 + 215

But we need to compute P'(x) = 0:

-x/15 + 215 = 0

x/15 = 215

x = 3225

Also the second derivative of P(x) is given as:

P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3225. Hence:

P(3225) =  -(3225)²/30 + 215(3225) - 74000 = $272687.5

The money to be charge for each set is:

p(x) = 300 - 3225/30 = $192.5 per set

When taxed $5, the maximum profit is $272687.5

3 0
2 years ago
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