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murzikaleks [220]
3 years ago
8

Can someone help me with this I really don't understand​

Mathematics
2 answers:
castortr0y [4]3 years ago
6 0
It’s increasing by 1,2 so keep going by 1,2 and see if u get to 20,13 if not it means it’s not on that line
Andrej [43]3 years ago
6 0

Answer:

No

Step-by-step explanation:

Because the constant of proportionality is not equivalent to 20,13.

Can I have brainliest?

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The equation is as follows

c = 16h
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A recipe for rice and lentils calls for 1 L of broth for every 250 g of lentils
vladimir2022 [97]
625 i think but im just multiplying by 2.5
8 0
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A square has an area of 64 ft? What is the length of each side?
exis [7]

Answer:

8 ft.

Step-by-step explanation:

The formula for area of a square is:

A=s^2

We are given the area of 64 square feet.

64=s^2\\\\\sqrt{64}=\sqrt{s^2}\\\\  \boxed{8=s}

Each side should be 8 ft.

Hope this helps.

3 0
4 years ago
A - H, please! thank you.
liberstina [14]

Answer:

First exercise

a) 1/8=0.125

b) 5$

c) y = 0.125 * x - 5

Second exercise

a) 0.11 $/KWh

b) no flat fee

c) y = 0.11 * x

Step-by-step explanation:

(See the pictures)

4 0
3 years ago
what are the x-coordinates for the maximum points in the function f(x) = 4 cos(2x- pi) from x = 0 to x= 2 pi?
julsineya [31]

Answer:

\frac{\pi }{2} and \frac{3\pi }{2}

Step-by-step explanation:

To find the max points we need to take the derivative of the function and then find the critical values.

First we take the derivative:

f(x) = 4cos(2x-\pi )\\f'(x)=-4sin(2x-\pi )(2)\\f'(x)=-8sin(2x-\pi )\\

Now we need to find when f'(x)=0 to find the critical values.

0=-8sin(2x-\pi )\\0=sin(2x-\pi )\\sin^{-1}0=2x-\pi \\0=2x-\pi \\\pi =2x\\\frac{\pi }{2} =x

The critical values will be

\frac{\pi }{2} n for any integer n

between 0 and 2 pi, the critical values will be

0, \frac{\pi }{2} ,\pi ,\frac{3\pi }{2},2\pi

We can determine if these are minimums or maximums by using the second derivative test.

So we need to take the second derivative;

f'(x)=-8sin(2x-\pi )\\f''(x) = -8cos(2x-\pi )(2)\\f''(x)=-16cos(2x-\pi)

We need to see if the second derivative is positive or negative to determine if it is a max or min.

f''(0) = 16\\f''(\frac{\pi}{2})=-16\\f''(\pi )=16\\f''(\frac{2\pi}{3}) = -16\\

Since the second derivative is negative at

\frac{\pi }{2} and \frac{3\pi }{2}

we know both of those are the x-values of maximums.

5 0
4 years ago
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