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Natalija [7]
3 years ago
12

Part K Whose sequence proves triangle ABC is congruent to triangle DEF? Why?

Mathematics
1 answer:
densk [106]3 years ago
4 0
The answer : It rotates around
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stepan [7]
I’m pretty sure it’s Dish A. They’re both starting with 5 clusters, but the graph of Dish B doesn’t show new growth.










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3 years ago
I need help with this question!! Show steps please
Butoxors [25]
Parallelogram: Both sides are of similar length and are parallel to one another.

Since the length of both sides need to be the same,
Length of one side:
3-(-4) = 7

Length of other side = 7
y coordinate of fourth vertex = -2+7 = 5

Since both lines are parallel, the x coordinate of the fourth vertex is similar to the point below it, which is -4

Therefore, the coordinate of the fourth vertex is (-4, 5)
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3 years ago
- 10 + x < - 2 what is the solution of the inequality shown below
Mama L [17]

Answer:

x<8

Step-by-step explanation:

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3 years ago
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b. 41.43gallons

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6 0
3 years ago
The expression (1/50)(sqrt(1/50)+sqrt(2/50)+sqrt(3/50)+...+sqrt(50/50) is a Riemann Sum approximation for
alexira [117]
\dfrac1{50}\left(\sqrt{\dfrac1{50}}+\sqrt{\dfrac2{50}}+\cdots+\sqrt{\dfrac{50}{50}}\right)=\displaystyle\sum_{n=1}^{50}\sqrt{\dfrac n{50}}\frac1{50}

describes a sum of the areas of 50 rectangles, each of width \dfrac1{50}, and the nth rectangle has height \sqrt{\dfrac n{50}}.

\sqrt{\dfrac n{50}} ranges from a small positive number to 1, which means the integration interval must be [0,1]. The sum is then the right-endpoint Riemann sum of \sqrt x over the interval with 50 equally spaced subintervals, so

\displaystyle\sum_{n=1}^{50}\sqrt{\dfrac n{50}}\frac1{50}\approx\int_0^1\sqrt x\,\mathrm dx

The sum itself evaluates to roughly 0.676, while the exact value of the integral is \dfrac23\approx0.667.
3 0
3 years ago
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