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mars1129 [50]
3 years ago
9

Answer that for me please i am very confused

Mathematics
2 answers:
madreJ [45]3 years ago
4 0
They are asking how to rewrite the equation with it as x= what ever comes after
example 4*x=20  so x= 20/4
Alex787 [66]3 years ago
4 0
The answer is A f(X)=X+4  
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I think the distance is 7

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3 years ago
Lim x→π/2 1-sinx/cot^2x<br>any genious help please ​
Simora [160]

Rewrite the limand as

(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = (1 - sin(<em>x</em>)) / (cos²(<em>x</em>) / sin²(<em>x</em>))

… = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / cos²(<em>x</em>)

Recall the Pythagorean identity,

sin²(<em>x</em>) + cos²(<em>x</em>) = 1

Then

(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / (1 - sin²(<em>x</em>))

Factorize the denominator; it's a difference of squares, so

1 - sin²(<em>x</em>) = (1 - sin(<em>x</em>)) (1 + sin(<em>x</em>))

Cancel the common factor of 1 - sin(<em>x</em>) in the numerator and denominator:

(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = sin²(<em>x</em>) / (1 + sin(<em>x</em>))

Now the limand is continuous at <em>x</em> = <em>π</em>/2, so

\displaystyle\lim_{x\to\frac\pi2}\frac{1-\sin(x)}{\cot^2(x)}=\lim_{x\to\frac\pi2}\frac{\sin^2(x)}{1+\sin(x)}=\frac{\sin^2\left(\frac\pi2\right)}{1+\sin\left(\frac\pi2\right)}=\boxed{\frac12}

4 0
3 years ago
What is the value of the nearest tenth ?
JulsSmile [24]

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1

Step-by-step explanation:

3 0
3 years ago
PLZ HELP I WILL MARK THE BRAINIEST slove for x on each problem
dem82 [27]

Answer:

1-141

2-39

3-139

4-98

5-102

6-50

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9-95

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Step-by-step explanation:

6 0
4 years ago
A piece of wire is 76cm long. It is cut into two unequal parts and each part is bent into a square. The sum of the areas of the
NemiM [27]

Answer:

The length of the shorter part of the wire is 24 centimeters.

Step-by-step explanation:

Let L the total length of the piece of wire, where L-x and x are the perimeters of the greater and lesser squares. All lengths are measured in centimeters. Since squares have four sides of equal length, the side lengths for the greater and lesser squares are \frac{L-x}{4} and \frac{x}{4}. From statement we find that the sum of the areas of the two squares (A), measured in square centimeters, is represented by the following expression:

A = \left(\frac{L-x}{4} \right)^{2}+\left(\frac{x}{4} \right)^{2} (1)

And we expand this polynomial below:

A = \frac{L^{2}-2\cdot L\cdot x +x^{2}}{16} + \frac{x^{2}}{16}

A = \frac{L^{2}-2\cdot L\cdot x +2\cdot x^{2}}{16}

2\cdot x^{2}-2\cdot L\cdot x +L^{2}-16\cdot A = 0 (2)

If we know that L = 76\,cm and A = 205\,cm^{2 }, then the length of the shorter part of the wire is:

2\cdot x^{2}-152\cdot x +2496 = 0

By the Quadratic Formula, we determine the roots associated with the polynomial:

x_{1} = 52\,cm, x_{2} = 24\,cm

The length of the shorter part of the wire corresponds to the second root. Hence, the length of the shorter part of the wire is 24 centimeters.

3 0
3 years ago
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