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3 years ago

What is 6.43 rounded to the nearest tenth

2 answers:

7 0

6.43 rounded to the nearest tenth is 6.40

The place after the decimal is the tenths place. We rely on the value of the hundredths place to know whether we round it up or we round it down to the nearest tenths. If the number is 5 and above, we round it up. If the number is below 5, we round it down.

3 is below 5 therefore from 6.43 we round it down to 6.40

xeze [42]3 years ago

5 0

6.43 rounded to the nearest tenth is 6.4.

This is because 6.43 is closer to 6.4 than it is to 6.5.

Your final answer is 6.4.

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PLEASE HELP ILL GIVE MEDALS AND MARK BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!! NEEDS TO BE ALEGABRA 2 MEATHOD!!!!!!!!

nevsk [136]

Simplify \frac{5}{3}x35x to \frac{5x}{3}35x

x-\frac{5x}{3}<3x−35x<3

2

Simplify x-\frac{5x}{3}x−35x to -\frac{2x}{3}−32x

-\frac{2x}{3}<3−32x<3

3

Multiply both sides by 33

-2x<3\times 3−2x<3×3

4

Simplify 3\times 33×3 to 99

-2x<9−2x<9

5

Divide both sides by -2−2

x>-\frac{9}{2}x>−29

8 0

3 years ago

1. Is triangle P’Q’R larger or smaller than triangle PQR? Explain how you know

Salsk061 [2.6K]

Answer:

The length of the line is PQ as this line is parallel to the x - axis. So, the length of the line is the summation of 10 from second quadrant and 20 from first quadrant. So, the sum is 30. Hence the length of the line is 30 units.

Step-by-step explanation:

The length of a line segment can be measured by measuring the distance between its two endpoints. It is the path between the two points with a definite length that can be measured. Explanation: On a graph, the length of a line segment can be found by using the distance formula between its endpoints.

7 0

2 years ago

Please help me answer one of these!

babunello [35]

This seems to be referring to a particular construction of the perpendicular bisector of a segment which is not shown. Typically we set our compass needle on one endpoint of the segment and compass pencil on the other and draw the circle, and then swap endpoints and draw the other circle, then the line through the intersections of the circles is the perpendicular bisector.

There aren't any parallel lines involved in the above described construction, so I'll skip the first one.

2. Why do the circles have to be congruent ...

The perpendicular bisector is the set of points equidistant from the two endpoints of the segment. Constructing two circles of the same radius, centered on each endpoint, guarantees that the places they meet will be the same distance from both endpoints. If the radii were different the meets wouldn't be equidistant from the endpoints so wouldn't be on the perpendicular bisector.

3. ... circles of different sizes ...

[We just answered that. Let's do it again.]

Let's say we have a circle centered on each endpoint with different radii. Any point where the two circles meet will then be a different distance from one endpoint of the segment than from the other. Since the perpendicular bisector is the points that are the same distance from each endpoint, the intersection of circles with different radii isn't on it.

4. ... construct the perpendicular bisector ... a different way?

Maybe what I first described is different; there are no parallel lines.

8 0

3 years ago

Find the area of this parallelogram.

Alex17521 [72]

Answer:

it will be c

Step-by-step explanation:

8 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%

xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

$\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }$

Let a = 1 and b the cosine product, and write them as

$\dfrac{a - b}{x^2}$

with

$b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}$

Now we use the identity

$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$

to rationalize the numerator. This gives

$\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}$

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

$\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}$

For the remaining limit, use the Taylor expansion for cos(x) :

$\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)$

where $\mathcal{O}(x^4)$ essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2$

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)$

$\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)$

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

$\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52$

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0

2 years ago