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3 years ago

Given that PO is a midsegment of △LMN, to which segment is MO congruent?

Mathematics

2 answers:

Debora [2.8K]3 years ago

7 0

Answer:just got it right!

Step-by-step explanation:

Paladinen [302]3 years ago

5 0

Answer:

2nd option, OL

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Determine whether the relationship is a function or not. Explain why or why not!

Kruka [31]

Answer:

Yes or No: Yes

Explanation: It is a function because each input is paired with only one output, as shown in the table.

Step-by-step explanation:

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3 years ago

A spinner has 4 equal-sized sections. To win the game, the pointer must land on a purple section.

expeople1 [14]

Answer: 1/4

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4 years ago

If FG = 2 units, FI = 7 units, and HI = 1 unit, what is GH? 3 units 4 units 5 units 6 units

Nezavi [6.7K]

I think the points given here are plotted linearly:

FGHI. in this case, we can tell that FG + GH + HI = FI. substituting to the expression devised, 2 units + GH + 1 unit = 7 units. This is equal to 3 units + GH = 7 units. GH is then equal to 4 units.

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4 years ago

Read 2 more answers

What is the area of the composite figure? 65 cm2 52.5 cm2 40 cm2 60 cm2

Sergio [31]

Answer:

65 cm²

Step-by-step explanation:

follow for more on my id

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3 years ago

Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ

SOVA2 [1]

$y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}$

The arc length of the curve is

$\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx$

which has a value of about 5.99086.

Let $f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}$ . Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

$\ell_i=\dfrac{i-1}2$

$r_i=\dfrac i2$

with $1\le i\le10$ .

These subintervals have midpoints given by

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

Over each subinterval, we approximate $f(x)$ with the quadratic polynomial

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that the integral we want to find can be estimated as

$\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that

$\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6$

so that the arc length is approximately

$\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086$

5 0

3 years ago