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ch4aika [34]
3 years ago
10

Evaluate, when x=2 y=-5 and z=3

Mathematics
1 answer:
gregori [183]3 years ago
8 0

Answer:

-139

Step-by-step explanation:

Evaluate 1/4 (4 x^3 - 2 y - 2 z^3) y^2 - 16 x^2 where x = 2, y = -5 and z = 3:

(4 x^3 - 2 y - 2 z^3)/4 y^2 - 16 x^2 = (4×2^3 - -5×2 - 2×3^3)/4×(-5)^2 - 16×2^2

(4×2^3 - 2 (-5) - 2×3^3)/4×(-5)^2 = ((4×2^3 - 2 (-5) - 2×3^3) (-5)^2)/4:

((4×2^3 - 2 (-5) - 2×3^3) (-5)^2)/4 - 16×2^2

(-5)^2 = 25:

((4×2^3 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

2^3 = 2×2^2:

((4×2×2^2 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

2^2 = 4:

((4×2×4 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

2×4 = 8:

((4×8 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

3^3 = 3×3^2:

((4×8 - 2 (-5) - 23×3^2) 25)/4 - 16×2^2

3^2 = 9:

((4×8 - 2 (-5) - 2×3×9) 25)/4 - 16×2^2

3×9 = 27:

((4×8 - 2 (-5) - 227) 25)/4 - 16×2^2

4×8 = 32:

((32 - 2 (-5) - 2×27) 25)/4 - 16×2^2

-2 (-5) = 10:

((32 + 10 - 2×27) 25)/4 - 16×2^2

-2×27 = -54:

((32 + 10 + -54) 25)/4 - 16×2^2

| 3 | 2

+ | 1 | 0

| 4 | 2:

(42 - 54 25)/4 - 16×2^2

42 - 54 = -(54 - 42):

(-(54 - 42) 25)/4 - 16×2^2

| 5 | 4

- | 4 | 2

| 1 | 2:

(-12×25)/4 - 16×2^2

(-12)/4 = (4 (-3))/4 = -3:

-3×25 - 16×2^2

2^2 = 4:

-3×25 - 164

-3×25 = -75:

-75 - 16×4

-16×4 = -64:

-64 - 75

-75 - 64 = -(75 + 64):

-(75 + 64)

| 7 | 5

+ | 6 | 4

1 | 3 | 9:

Answer:  -139

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Use the denition of the derivative to find f 0 (3), where f (x) = 3x+5 / 2x−1
Crazy boy [7]

Answer:

f'(3)= -\frac{13}{25}

Step-by-step explanation:

We are asked to find f'(3) of function f(x)=\frac{3x+5}{2x-1} using definition of derivatives.

Limit definition of derivatives:

f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

Let us find f(3+h) and f(3).

f(3+h)=\frac{3(3+h)+5}{2(3+h)-1}

f(3+h)=\frac{9+3h+5}{6+2h-1}\\\\f(3+h)=\frac{3h+14}{2h+5}

f(3)=\frac{3(3)+5}{2(3)-1}

f(3)=\frac{9+5}{6-1}\\\\f(3)=\frac{14}{5}

Substituting these values in limit definition of derivatives, we will get:

f'(3)= \lim_{h \to 0} \frac{f(3+h)-f(3)}{h}

f'(3)= \lim_{h \to 0} \frac{\frac{3h+14}{2h+5}-\frac{14}{5}}{h}

Make a common denominator:

f'(3)= \lim_{h \to 0} \frac{\frac{(3h+14)*5}{(2h+5)*5}-\frac{14(2h+5)}{5(2h+5)}}{h}

f'(3)= \lim_{h \to 0} \frac{\frac{5(3h+14)-14(2h+5)}{5(2h+5)}}{h}

f'(3)= \lim_{h \to 0} \frac{5(3h+14)-14(2h+5)}{5h(2h+5)}

f'(3)= \lim_{h \to 0} \frac{15h+70-28h-70}{5h(2h+5)}

f'(3)= \lim_{h \to 0} \frac{-13h}{5h(2h+5)}

Cancel out h:

f'(3)= \lim_{h \to 0} \frac{-13}{5(2h+5)}

f'(3)= \frac{-13}{5(2(0)+5)}

f'(3)= \frac{-13}{5(5)}

f'(3)= -\frac{13}{25}

Therefore, f'(3)= -\frac{13}{25}.

8 0
3 years ago
What is the least number that has 4 odd factors that are all the same? each factor is greater than 1, and can have only 1 and it
Yakvenalex [24]

Answer:

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Step-by-step explanation:

We need to have 4 odd factors not including 1 that are the same

Listing the odd numbers

1,3,5,7,9,11,13,.....

3 is the smallest odd number that is not 1

The factors of 3 are 1 and 3

The number is 3*3*3*3

                       81

7 0
2 years ago
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