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valentina_108 [34]
2 years ago
13

Someoone plz help me

Mathematics
1 answer:
posledela2 years ago
7 0
The ratio depicts that their is 63 boys and 16 girls based on the ratio 7:4.
I hope this helps!
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I need the domain range and function. With an explanation
erastovalidia [21]
<h3>Answers:</h3>

Problem 1

  • Domain = -3 < x \le 3, interval notation (-3, 3]
  • Range = -3 \le y < 3, interval notation [-3, 3)
  • Is it a function? Yes

Problem 2

  • Domain = x \ge -2, interval notation [-2, \infty)
  • Range = All real numbers, interval notation (-\infty, \infty)
  • Is it a function? No

Problem 3

  • Domain = -4 \le x < 3, interval notation [-4, 3)
  • Range = -4 < y \le 3, interval notation (-4, 3]
  • Is it a function? Yes

Problem 4

  • Domain = All real numbers, interval notation (-\infty, \infty)
  • Range = y \le 4, interval notation (-\infty, 4]
  • Is it a function? Yes

==================================================

Explanations:

  1. The left most point is when x = -3, and we are not including this value due to the open hole. The other endpoint is included because it is a filled in circle. The domain is therefore -3 < x \le 3 which in interval notation is (-3, 3]. We have the curved parenthesis meaning "exclude endpoint" and the square bracket says "include endpoint". The range is a similar story but we're looking at the smallest and largest y values. Though be careful about which endpoint is open/closed. We have a function because it passes the vertical line test.
  2. The smallest x value is x = -2. There is no largest x value because the arrows say to go on forever to the right. We can say the domain is x \ge -2 which in interval notation is [-2, \infty). The range is (-\infty, \infty) to indicate "all real numbers". This graph fails the vertical line test, so it is not a function. The vertical line test is where we check to see if we can pass a vertical line through more than one point on the curve. In this case, such a thing is possible which is why it fails the test.
  3. This is the same idea as problem 1, though note the endpoints are flipped in terms of which has an open circle and which doesn't. It is not possible to draw a single vertical line to have it pass through more than one point on the curve, so it passes the vertical line test and we have a function.
  4. This is a function because it passes the vertical line test. The domain is the set of all real numbers due to the arrows in both directions. Any x value is a possible input. The range is y \le 4 which is the same as saying (-\infty, 4] in interval notation. This is because y = 4 is the largest y value possible. There is no smallest y value due to the arrows.

7 0
3 years ago
Which of the following is not a requirement of a standard form equation Ax+By=C? a)A ≥ 0. b)B ≥ 0. c)A and B are not both 0. d)A
Ghella [55]
<h2>Answer:</h2>

The following which is not a requirement of a standard form of equation Ax+By=C is:

        b)   B ≥ 0    

<h2>Step-by-step explanation:</h2>

We know that the standard equation of a line is given by:

        Ax+By=C

where A,B and C are integers and A is taken to be a non-negative integer i.e. (A≥0) also the greatest common factor of A,B and C is: 1.

Also A and B can't be both zero.

       Hence, the correct option is:

                Option: b

7 0
2 years ago
Read 2 more answers
The weight of two bags is 24/15kg . If the weight of one bag is 10/15. What will be the weight of other bag?
lys-0071 [83]
Weight of other bag would be 14/15 kg
5 0
3 years ago
Find the minimum value of
vekshin1

Step-by-step explanation:

C=y[20]

Cy=[20] Answer is Cy=[20]

3 0
2 years ago
What Val are restricted from the domain of
djverab [1.8K]

<u>Given</u>:

The given expression is \frac{\left(4 x^{2}-4 x+1\right)}{(2 x-1)^{2}}

We need to determine the values for which the domain is restricted.

<u>Restricted values:</u>

Let us determine the values restricted from the domain.

To determine the restricted values from the domain, let us set the denominator the function not equal to zero.

Thus, we have;

(2x-1)^2\neq 0

Taking square root on both sides, we get;

2x-1\neq 0

     2x\neq 1

      x\neq \frac{1}{2}

Thus, the restricted value from the domain is x\neq \frac{1}{2}

Hence, Option A is the correct answer.

7 0
3 years ago
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