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AURORKA [14]
3 years ago
8

A college student needs 11 classes that are worth a total of 40 credits in order to complete her degree. The college offers both

4-credit classes and 3-credit classes. How many 4-credit classes does this college student need to complete her degree?
Mathematics
1 answer:
ipn [44]3 years ago
3 0
7 4-credit classes. 4 3-credit classes.

7*4 = 28, and 4*3 = 12. 11 classes, and 40 credits.
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If 32x+1 - 3x+5, what is the value of x?<br> оооо
Zolol [24]

Answer:

x = \frac{4}{35}

Step-by-step explanation:

Given

32x + 1 = - 3x + 5 ( add 3x to both sides )

35x + 1 = 5 ( subtract 1 from both sides )

35x = 4 ( divide both sides by 35 )

x = \frac{4}{35}

7 0
3 years ago
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Helpp me pleasee (algebra)
antiseptic1488 [7]

Answer:

Step-by-step explanation:

answer = 1/2 r + m

m = 7

r = 8

answer = 1/2 * 8 + 7

answer = 4 + 7

answer = 11

8 0
3 years ago
50+ POINTS I NEED HELP
olga nikolaevna [1]

Answer:

<u><em>21 yard line</em></u>

Step-by-step explanation:

Ok we start out on the 20 yard line, we need to add the yards together.

<em>We add the 6 to the 20:</em>

20+6=26

<em>Then we subtract the 8 yards from our new 26:</em>

26-8=18

<em>Then we add the 18 by 3:</em>

18+3=21

Therefore,

<em>The Answer is 21 Yard Line.</em>

Hope I could help, If you need some more assistance ask me!

8 0
3 years ago
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Over [174]

Answer:

15.25×3

45.75-5

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3 0
3 years ago
Find the critical numbers of the function f(x) = x6(x − 1)5 what does the first derivative test tell you that the second derivat
patriot [66]
The given function is f(x) = x⁶(x-1)⁵

The first derivative is
f'(x) = 6x⁵(x-1)⁵ + 5x⁶(x-1)⁴
       = x⁵(x-1)⁴(6x - 6 + 5x)
       = x⁵(x-1)⁴(11x - 6)
The critical values are the zeros of f'(x). They are
x = 0, 6/11, and 1.
The critical values indicate that turning points exist for f(x) at the critical points. However, we do not know the nature of the turning points.

Write the first derivative in the form f'(x) = (x-1)⁴(11x⁶ - 6x⁵).
The second derivative is
f''(x) = 4(x-1)³(11x⁶ - 6x⁵) + (x-1)⁴(66x⁵ - 30x⁴)
        = (x-1)³(44x⁶ - 24x⁵ + 66x⁶ - 30x⁵ - 66x⁵ + 30x⁴)
        = (x-1)³(110x⁶ - 120x⁵ + 30x⁴)
        = 10x⁴(x-1)³(11x² - 12x + 3)
The sign of f''(x) at the critical values tell us the nature of the turning point.

f''(0) = 0, therefore a point of inflection exists at x = 0.
f''(6/11) > 0, therefore a local minumum exists at x = 6/11.
f''(1) = 0, therefore a point of inflection exists at x=1.

The graphs shown below confirm these results.

8 0
3 years ago
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