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bulgar [2K]
2 years ago
11

What is the domain and range for following graph

Mathematics
1 answer:
Alex_Xolod [135]2 years ago
4 0

Answer:

the domain is -5 to 5; the range is -2.25 to 2.25

Step-by-step explanation:

the graph has no marking on it, so I suppose each block represents 1 unit

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See screenshots for question and answer choices.
yawa3891 [41]

Answer:

I belive its 300

Step-by-step explanation:

Brainlist?

3 0
3 years ago
Read 2 more answers
If xy + y² = 6, then the value of dy/dx at x= -1 is<br>​
mylen [45]

Hi there!

\large\boxed{\text{At (-1, -2), }\frac{dy}{dx} = -\frac{2}{5}}}

\large\boxed{\text{At (-1, 3), }\frac{dy}{dx} = -\frac{3}{5}}}

We can calculate dy/dx using implicit differentiation:

xy + y² = 6

Differentiate both sides. Remember to use the Product Rule for the "xy" term:

(1)y + x(dy/dx)  + 2y(dy/dx) = 0

Move y to the opposite side:

x(dy/dx) + 2y(dy/dx) = -y

Factor out dy/dx:

dy/dx(x + 2y) = -y

Divide both sides by x + 2y:

dy/dx = -y/x + 2y

We need both x and y to find dy/dx, so plug in the given value of x into the original equation:

-1(y) + y² = 6

-y + y² = 6

y² - y - 6 = 0

(y - 3)(y + 2) = 0

Thus, y = -2 and 3.

We can calculate dy/dx at each point:

At y = -2: dy/dx = -(-2) / -1+ 2(-2) = -2/5.

At y = 3: dy/dx = -(3) / -1 + 2(3) = -3/5.

5 0
3 years ago
What is the best order-of-magnitude for 0.000006?
seraphim [82]
So,

Optimally, we want a number between 1 and 10 multiplied by a power of ten.

If we move the decimal place 6 places to the right, we get 6.

So we want 10^{-6}.

The correct option is C.
7 0
3 years ago
HELP PLS! whoever gets it right gets brainliest!!
galben [10]

Answer: Pretty sure it is 3

Step-by-step explanation:

I think...

7 0
2 years ago
Read 2 more answers
George earn 3 dollars each time he sweeps and mops in the family room for you trying to earn $17 for tickets to the end of the s
uranmaximum [27]
17+4=21 
21 divided by 3= 7
17 dollars for tickets 
21-4=17
so your answer is 21 he has to sweep the floor 21 times to get tickets

5 0
3 years ago
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