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elixir [45]
3 years ago
9

Can i get help with these two questions please

Mathematics
1 answer:
Trava [24]3 years ago
4 0

Answer:

Step-by-step explanation:

i think you will replace x and y respectively,as in take 1\3 as x and -7\2 as y and so on as the next question

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26 cm

Step-by-step explanation:

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y=3x+1

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Find the minimum and maximum value of the function on the given interval by comparing values at the critical points and endpoint
Kitty [74]

Answer:

maximum: y = 1

minimum: y = 0.

Step-by-step explanation:

Here we have the function:

y = f(x) =  √(1 + x^2 - 2x)

we want to find the minimum and maximum in the segment [0, 1]

First, we evaluate in the endpoints, which are 0 and 1.

f(0)  =√(1 + 0^2 - 2*0) = 1

f(1) = √(1 + 1^2 - 2*1) = 0

Now let's look at the critical points (the zeros of the first derivate)

To derivate our function, we can use the chain rule:

f(x) = h(g(x))

then

f'(x) = h'(g(x))*g(x)

Here we can define:

h(x) = √x

g(x) = 1 + x^2 - 2x

Then:

f(x) = h(g(x))

f'(x)  =  1/2*( 1 + x^2 - 2x)*(2x - 2)

f'(x) = (1 + x^2 - 2x)*(x - 1)

f'(x) = x^3 - 3x^2 + x - 1

this function does not have any zero in the segment [0, 1] (you can look it in the image below)

Thus, the function does not have critical points in the segment.

Then the maximum and minimum are given by the endpoints.

The maximum is 1 (when x = 0)

the minimum is 0 (when x = 1)

7 0
3 years ago
If tan y = 3 over y and cos x = y over z what is the value of sin x??
LekaFEV [45]
The answer should be C
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3 years ago
If there are 6 serving in a 2/3 pound lb package of pound is in each serving
Serhud [2]

Answer:

\boxed{\math{\frac{1}{9}\text{ lb}}}

Step-by-step explanation:

\text{Size of one serving} = \dfrac{\text{Total size}}{\text{No of servings}} = \dfrac{\frac{2}{3}\text{ lb}}{\text{6 servings}}\\\\\text{Change the 6 to a fraction and change divide to multiply}\\\\\dfrac{2}{3} \div 6 = \dfrac{2}{3} \times \dfrac{1}{6}

\text{Cancel the 2s}\\\\\dfrac{2}{3} \times \dfrac{1}{6} = \dfrac{1}{3} \times \dfrac{1}{3}\\\\\text{Multiply numerators and denominators}\\\\\dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{9}\\\\\text{Each serving contains }\boxed{\mathbf{\frac{1}{9}\textbf{ lb}}}

3 0
3 years ago
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