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ki77a [65]
2 years ago
15

Use your table to work out what fraction of the fish are 50cm or more in length

Mathematics
2 answers:
CaHeK987 [17]2 years ago
8 0

Answer:

very very

Step-by-step explanation:

⟼sinA−cosAsinA+cosA+sinA+cosAsinA−cosA

Let,

sinA be a

cosA be b

⟼ \frac{a + b}{a - b} + \frac{a - b}{a + b}⟼a−ba+b+a+ba−b

⟼ \frac{(a + b)^{2} + (a - b) ^{2}}{(a + b)(a - b)}⟼(a+b)(a−b)(a+b)2+(a−b)2

⟼ \frac{2( {a}^{2} \: {b}^{2} ) }{ {a}^{2} - {b}^{2} }⟼a2−b22(a2b2)

Putting values

⟼ \frac{2 \sin^{2} A \: + { \cos }^{2} A }{ { \sin }^{2}A- { \cos }^{2}A }⟼sin2A−cos2A2sin2A+cos2A

⟼ \frac{2}{ { \sin}^{2} A - \cos^{2}A}⟼sin2A−cos2A2

Hope it helps you!!

#IndianMurga

antiseptic1488 [7]2 years ago
8 0

We need to find frequency in the region.

\\ \sf\longmapsto 50\leqslant l

We should calculate in it frequencies of below

\\ \sf\longmapsto 50\leqslant l

\\ \sf\longmapsto 60\leqslant l

\\ \sf\longmapsto 80\leqslant l

\\ \sf\longmapsto 100\leqslant l

Hence

total frequency:-

\\ \sf\longmapsto 8+b+c+d

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kondaur [170]
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the ordered pair represents 10 pounds of cabbage for 6 dollars
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2 years ago
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At a tailor shop, it costs $6.79 to
Margarita [4]

Answer: $29.62

Step-by-step explanation:

6.79 * 3 = 20.37

9.25 * 1 = 9.25

So, 20.37 + 9.25 = 29.62

7 0
2 years ago
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Please help me or tell me the answer
arsen [322]
B and F because they would be multiplied by 100, which would move the digits two times to the right.
7 0
3 years ago
The area of a rectangular community the garden is 300 square feet. The length of a rectang!
FromTheMoon [43]

Answer:

30, 10

Step-by-step explanation:

length=x

wdith=y

xy=300, x=3y

replace x with 3y in the first equation to get

3y^2=300

y^2=100

since you cant have a negative side of a garden, y=10. x=30.

7 0
3 years ago
f(x) = 2<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class="latex
loris [4]

Answer:

No answer is possible

Step-by-step explanation:

First, we can identify what the parabola looks like.

A parabola of form ax²+bx+c opens upward if a > 0 and downward if a < 0. The a is what the x² is multiplied by, and in this case, it is positive 2. Therefore, this parabola opens upward.

Next, the vertex of a parabola is equal to -b/(2a). Here, b (what x is multiplied by) is 1 and a =2, so -b/(2a) = -1/4 = -0.25.

This means that the parabola opens upward, and is going down until it reaches the vertex of x=-0.25 and up after that point. Graphing the function confirms this.

Given these, we can then solve for when the endpoints of the interval are reached and go from there.

The first endpoint in -2 ≤ f(x) ≤ 16 is f(x) = 2. Therefore, we can solve for f(x)=-2 by saying

2x²+x-4 = -2

add 2 to both sides to put everything on one side into a quadratic formula

2x²+x-2 = 0

To factor this, we first can identify, in ax²+bx+c, that a=2, b=1, and c=-2. We must find two values that add up to b=1 and multiply to c*a = -2  * 2 = -4. As (2,-2), (4,-1), and (-1,4) are the only integer values that multiply to -4, this will not work. We must apply the quadratic formula, so

x= (-b ± √(b²-4ac))/(2a)

x = (-1 ± √(1-(-4*2*2)))/(2*2)

= (-1 ± √(1+16))/4

= (-1 ± √17) / 4

when f(x) = -2

Next, we can solve for when f(x) = 16

2x²+x-4 = 16

subtract 16 from both sides to make this a quadratic equation

2x²+x-20 = 0

To factor, we must find two values that multiply to -40 and add up to 1. Nothing seems to work here in terms of whole numbers, so we can apply the quadratic formula, so

x = (-1 ± √(1-(-20*2*4)))/(2*2)

= (-1 ± √(1+160))/4

= (-1 ± √161)/4

Our two values of f(x) = -2 are (-1 ± √17) / 4 and our two values of f(x) = 16 are (-1 ± √161)/4 . Our vertex is at x=-0.25, so all values less than that are going down and all values greater than that are going up. We can notice that

(-1 - √17)/4 ≈ -1.3 and (-1-√161)/4 ≈ -3.4 are less than that value, while (-1+√17)/4 ≈ 0.8 and (-1+√161)/4 ≈ 2.9 are greater than that value. This means that when −2 ≤ f(x) ≤ 16 , we have two ranges -- from -3.4 to -1.3 and from 0.8 to 2.9 . Between -1.3 and 0.8, the function goes down then up, with all values less than f(x)=-2. Below -3.4 and above 2.9, all values are greater than f(x) = 16. One thing we can notice is that both ranges have a difference of approximately 2.1 between its high and low x values. The question asks for a value of a where a ≤ x ≤ a+3. As the difference between the high and low values are only 2.1, it would be impossible to have a range of greater than that.

7 0
2 years ago
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