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2 years ago

GEOMETRY) the question is on the picture below :)

Mathematics

1 answer:

aev [14]2 years ago

5 0

Answer: 33

Step-by-step explanation:

Angles inscribed in the same arc are congruent, so

$13+4x=-7+8x\\\\20+4x=8x\\\\20=4x\\\\x=5\\\\\implies m\angle CDE=-7+5(8)=33^{\circ}$

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What is the area of the figure?

AysviL [449]

Answer:

Step-by-step explanation:

3 0

3 years ago

The data below represents the number of essays that students in Mr. Ji's class wrote. 2,3,5,5,6,7,8,8,11 Which box plot correctl

sweet-ann [11.9K]

Answer:

Step-by-step explanation:

2,3,5,5,6,7,8,8,11

6 is the middle line, 8 is the end of the box, and 4 is the beginning of the bow.

Hope this helps.

5 0

3 years ago

Aneesha travels at a rate of 50 miles per hour. Morris is traveling 3 feet per second less than Aneesha. Which is the best estim

goldfiish [28.3K]

If Aneesha travels at a rate of
50 miles............................1h
and 1mile=5280ft, therefore 50 miles = 264000 ft
and 1 h =3600 seconds we can say that Aneesha travels

264,000 ft in 3600 seconds or
264,000/3600= 73.33 ft/s

Morris travels 3 ft per second less than her
so 73.33-3=70.33 ft/s for Morris

3 0

3 years ago

Last HW Mr Thompson! Please Help!

TiliK225 [7]

Answer:

B, D, D, B, C

Step-by-step explanation:

1) The part of the equation that tells us the graph would open downward would be B, the negative coefficient of $x^{2}$ .

We know this because, if we reorganize the equation into the standard form y = mx + b, we get y = -(x²) + 1. This means that for any value we substitute for x, y will always be negative. The answer is B.

2) To find the zeroes of this function, we just have to see which answer makes the function, well, zero.

We have the equation y = (-16t - 2) (t - 1)

We set the equation equal to zero like this:

0 = (-16t - 2) (t - 1)

We also know that anything multiplied by zero will equal zero, so:

0 = (-16t - 2) and 0 = (t - 1)

Now we just solve for t in both equations:

0 = -16t - 2

-16t = 2

t = 2/-16 or -1/8

0 = t - 1

t = 1

So our answer will be D.

3) This time, we are given the same equation but the variable t represents time and y represents the height.

The zeroes are when y = 0, so the zeroes (which are 1 and -1/8 from the last problem) mean that the height is also at zero (which is the ground).

Furthermore, time cannot be negative, so the correct answer choice would be D.

4) We are given the same equation and story problem, but instead we are asked to find the height when the ball was thrown. In other words, we need to find the y-intercept.

The y-intercept is when t = 0, so:

y = (-16t - 2) (t - 1)

y = (-16(0) - 2) (0 - 1)

y = (-2) (-1)

y = 2

So, the correct answer is B, 2 feet.

5) This problem is a simple theoretical question. We know that the constant c will represent the vertical translation or y-intercept, so the correct answer is C.

8 0

3 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