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3 years ago

13

Stephen uses 2/25 kilograms of tofu in each serving of his famous tofu dish. He has 1 1/10 kilograms of tofu. How many servings

can Stephen make? in fraction

1 answer:

larisa [96]3 years ago

7 0

Answer: He has 1 1/10 kilograms of tofu.

Step-by-step explanation: We will have to divide the number of tofu required with the number of tofu own. Stephen uses 2/25 kilograms of tofu in each serving and he has 11/10 kilograms tofu.

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Fofino [41]

Answer:

Left 4 units.

Step-by-step explanation:

g(x) = x^2 + 8x + 16 Factoring:

g(x) = (x + 4)^2.

So to obtain g(x) from f(x) = x^2, f(x) must be shifted 4 units to the left.

3 0

3 years ago

consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where

boyakko [2]

Answer:

a) $X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

b) $Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

Step-by-step explanation:

From the question we are told that

The Function

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

Generally the differentiation of function f(x) is mathematically solved as

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

$f(x)=\frac{x^3+x^2+5x+1}{x^2}$

Therefore

$f'(x)=\frac{x^2+10x+3}{x^4}$

Generally critical point is given as

$f'(x)=0$

$\frac{x^2+10x+3}{x^4}=0$

$x=-5 \pm\sqrt{22}$

Generally the maximum and minimum x value for critical point is mathematically solved as

$f'(-5 \pm\sqrt{22})$

Where

Maximum value of x

$f'(-5 +\sqrt{22})$

Minimum value of x

$f'(-5 +\sqrt{22})$

Therefore interval of increase is mathematically given by

$f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})$

$f(x)$

Therefore interval of decrease is mathematically given by

$(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)$

Generally the second differentiation of function f(x) is mathematically solved as

$f''(x)=\frac{2(x^2+15x+6)}{x^5}$

Generally the point of inflection is mathematically solved as

$f''(x)=0$

$x^2+15x+6=0$

Therefore inflection points is given as

$x=\frac{1}{2} (-15 \pm \sqrt{201}$

$f''(x)>0,\frac{1}{2}(-15-\sqrt{201})$

a)Generally the concave upward interval X is mathematically given as

$X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

$f''(x)$

b)Generally the concave downward interval Y is mathematically given as

$Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

5 0

3 years ago

9. Jack is learning to save, spend, and give his money away to charities. He has determined to give 10% of his summer job earnin

Kamila [148]

Answer:

$2560

Step-by-step explanation:

if 10%=256

then 10 times the amount 100% is 2560

4 0

3 years ago

Find the sum of the following series. Round to the nearest hundredth if necessary,

Aneli [31]

Answer:

$sum = \frac{a( {r}^{n - 1} )}{r - 1} \\ : but \: l = a( {r}^{n - 1} ) \\ 49152 = 3( {2}^{n - 1} ) \\ 16384 = {2}^{n - 1} \\ {2}^{n} = 32768 \\ {2}^{n} = {2}^{15} \\ n = 15 \\ \therefore \: sum = \frac{3(2 {}^{15 - 1}) }{15 - 1} \\ = \frac{49152}{14} \\ = 3510.9$

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3 years ago

Make y the subject<br> z+2 = 2/1-y

s344n2d4d5 [400]

Y= subject or y1/2=2

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2 years ago