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12345 [234]
2 years ago
13

The unit rate is .4 right

Mathematics
2 answers:
timofeeve [1]2 years ago
7 0

Answer:

yes

Step-by-step explanation:

LiRa [457]2 years ago
3 0

Answer:

Uh yes and no, im not sure

Step-by-step explanation:

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Hi, please help me out :)) thank you
r-ruslan [8.4K]

Answer:

d =32

Step-by-step explanation:

\sqrt{7d+1}-4=11\\\\\mathrm{Add\:}4\mathrm{\:to\:both\:sides}\\\sqrt{7d+1}-4+4=11+4\\\\Simplify\\\sqrt{7d+1}=15\\\\\mathrm{Square\:both\:sides}:\quad 7d+1=225\\7d+1=225\\\\\mathrm{Solve\:}\:7d+1=225:\\\quad d=32

4 0
3 years ago
Read 2 more answers
Plz answer urgent...A total of 247 students were surveyed about what they liked best for lunch. The results can be shown in a tw
adoni [48]

Answer:

Nein, du schlauer Penner

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
What is a third-degree binomial?
ankoles [38]

Answer:

  • B and D

Step-by-step explanation:

  • <em>Binomial = 2 terms in the expression</em>
  • <em>Third degree = highest added degree of variables in one term</em>

(A) It has 3 terms and highest degree of 4. It is trinomial of degree 4.

  • No

(B) It has 2 terms and highest degree of 3. It is binomial of degree 3.

  • Yes

(C) It has 2 terms and highest degree of 7. It is binomial of degree 7.

  • No

(D) It has 2 terms and highest degree of 3. It is binomial of degree 3.

  • Yes
8 0
3 years ago
Is the following true or false? d/dx[x^3e^x]=x^2e^x (3x+2)
wolverine [178]
It's false. It's a product so...
Derivative of the first TIMES the second PLUS derivative of second TIMES the first.

Derivative of the first (x^3) = 3x^2
Times the second = 3x^2 * e^x

Derivative of the second = e^x (remains unchanged)
Times the first = e^x * x^3
So the answer would be (3x^2)(e^x) + (e^x)(x^3)
which can be factorised to form x^2·e^x(3 + x)


3 0
3 years ago
Find the critical numbers and the intervals on which the function f(x)=9x5−3x3+6f(x)=9x5−3x3+6 is increasing or decreasing. Use
Ierofanga [76]

Answer:

x = (0, -√(3/5), √(3/5)) are the critical points.

Local maximum at √(3/5), and local minimum is at -√(3/5).

The function is increasing in the interval

(−∞, -√(3/5)) U (√(3/5), ∞)

And decreasing in the interval

(-√(3/5), 0) U (0, √(3/5))

Step-by-step explanation:

Given the function

f(x) = 9x^5 - 3x³ + 6

First of all, take the first derivative of this, to have

f'(x) = 45x^4 - 27x²

The critical point are the points where the first derivative vanishes, that is

f'(x) = 0

Now, solve the equation

45x^4 - 27x² = 0

9x²(5x² - 3) = 0

x = 0 twice

Or

5x² - 3 = 0

5x² = 3

x² = 3/5

x = ±√(3/5)

So, x = (0, -√(3/5), √(3/5)) are the critical points.

Local maximum is when f'(x) > 0, this is √(3/5) in this case,

and local minimum is when f'(x) < 0, this is -√(3/5) in this case.

Now, we need to test for the various intervals to determine where the function increases and decreases.

(−∞, -√(3/5)):

f'(-√(4/5)) = 45(-√(4/5))^4 - 27(-√(4/5))²

= 36/5 > 0. Increasing

(-√(3/5), 0):

f'(-√(2/5)) = 45(-√(2/5))^4 - 9(-√(2/5))²

= -18/5 < 0. Decreasing

(0, √(3/5)):

f'(√(1/5)) = 45(√(1/5))^4 - 9(√(1/5))² = -18/5 < 0. Decreasing

(√(3/5), ∞): f'(1) = 45(1)^4 - 9(1)² =

36 > 0. Increasing.

Therefore, the function is increasing in the interval

(−∞, -√(3/5)) U (√(3/5), ∞)

And decreasing in the interval

(-√(3/5), 0) U (0, √(3/5))

6 0
3 years ago
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