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hram777 [196]
3 years ago
10

Which of the two functions below has the largest maximum y-value?f(x) = -3x^4 - 14g(x) = -x^3 + 2 A.There is not enough informat

ion to determine B.g(x) C.f(x) D.The extreme maximum y-value for both f(x)and g(x) is infinity
Mathematics
2 answers:
Harrizon [31]3 years ago
8 0

Answer:

B is correct

Step-by-step explanation:

Naddik [55]3 years ago
4 0
We have the following functions:
 f (x) = -3x ^ 4 - 14 
 g (x) = -x ^ 3 + 2f
 Deriving both functions we have:
 f '(x) = -12x ^ 3
 g '(x) = -3x ^ 2
 We equal zero and clear x:
 f '(x) = -12x ^ 3 = 0 ----> x = 0
 g '(x) = -3x ^ 2 = 0 -----> x = 0
 Substituting x = 0 in the given equations:
 f (0) = -3 (0) ^ 4 - 14 = -14
 g (0) = - (0) ^ 3 + 2 = 2
 Answer:
 
the largest maximum y-value is:
 
B.g (x)
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4/5. To divide a fraction by a fraction you would flip the second fraction, simplify if possible and multiply like normal. 1/2 x 8/5 1/1 x 4/5 4/5 :)
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3 years ago
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irina [24]

Answer:

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11x = 180 - 5 - 4

11x = 171           and since theres mutliplication 11 times x which is 11x, were going to divide

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3 years ago
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Help finding the sum
Mamont248 [21]
It is asking you to find the sum of k^2 - 1 from k=1 to k=4. Since that is only 4 numbers, calculating the sum by hand wouldn’t be that bad.

(1^2 - 1) + (2^2 - 1) + (3^2 - 1) + (4^2 - 1) = 26

The easier way to find the sum is to use a few simple formulas.

When we have a term that is just a constant c, the formula is c*n.

When we have a variable k, the formula is k*n*(n+1)/2.

When we have a squared variable, the formula is k*n*(n+1)*(2n+1)/6.

In this case, we have a squared variable k^2 and a constant of -1.

So plug in n=4 to the formulas:

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5 0
3 years ago
a solar lease customer built up an excess of 6,500 kilowatts hour (kwh) during the summer using his solar panels. when he turned
Jlenok [28]

Question:

A solar lease customer built up an excess of 6,500 kilowatts hour (kwh) during the summer using his solar panels. when he turned his electric heat on, the excess be used up at 50 kilowatts hours per day .

(a) If E represents the excess left and d represent the number of days. Write an equation for E in terms of d

(b) How much of excess will be left after one month (1 month = 30 days)

Answer:

a. E = 6500 - 50d

b. E = 5000

Step-by-step explanation:

Given

Excess = 6500kwh

Rate = 50kwh/day

Solving (a): E in terms of d

The Excess left (E) in d days is calculated using:

E = Excess - Rate * days

The expression uses minus because there's a reduction in the excess kwh on a daily basis.

Substitute values for Excess, Rate and days

E = 6500 - 50 * d

E = 6500 - 50d

Solving (b); The value of E when d = 30.

Substitute 30 for d in E = 6500 - 50d

E = 6500 - 50 * 30

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4 0
3 years ago
I’m stuck please help ? Due tomorrow
Shkiper50 [21]
Here are two ways of doing it.

************************************************************

1) The current price is 100% of the price. The price went down by 6%, so since 100% - 6% = 94%, the new price is 94% of the original price.

94% * 175 = 0.94 * 175 = 164.50

The new price is 164.50

*******************************************************

2) The discount is 6% of 175, so first, we find the discount.

6% of 175 = 0.06 * 175 = 10.5

Now we subtract the amount of the discount from the original price.

175 - 10.50 = 164.50

The new price is 164.50

**********************************************************

As you can see, both methods give you the same answer, 164.50
6 0
3 years ago
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