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maxonik [38]
2 years ago
12

Label L, U, N, A with the correct angles. please help! (NO BOTS!)

Mathematics
2 answers:
Katyanochek1 [597]2 years ago
7 0

Answer:

L= 126 degrees, U= 66 degrees, N= 48 degrees, A= 54 degrees

Step-by-step explanation:

The magic number is 180, that is when all angles form a strait line. So, in order to solve for each angle, you have to add and subtract a little. Some of the angles are the same degree as others because they are equivalent to their opposite angle. Hope that helps :)

P.S. I took geometry over two years ago so I know my stuff lol, I'm 100% sure. Cheers!

OLEGan [10]2 years ago
5 0
The answer above it 100% correct I just had this question
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4x^2+16x+14=0
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3 0
3 years ago
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A 4-person task force is to be formed from the 4 men and 3 women who work in Company G’s human resources department. If there ar
Dima020 [189]

Answer:

Total 18 task force can be formed.

Step-by-step explanation:

Total number of men = 4

Total number of women = 3

We need to form a 4-person task force.

The total number of ways of selecting r items from n items is

^nC_r=\frac{n!}{r!(n-r)!}

Total number of ways of selecting 2 men from 4 men = ^4C_2

Total number of ways of selecting 2 women from 3 men = ^3C_2

Total number of different task forces that can be formed is

Total=^4C_2\times ^3C_2

Total=\frac{4!}{2!(4-2)!}\times \frac{3!}{2!(3-2)!}

Total=\frac{4!}{2!2!}\times \frac{3!}{2!1!}

Total=\frac{4\times 3\times 2!}{2\times 1\times 2!}\times \frac{3\tmes 2!}{2!}

Cancel out common factors.

Total=6\times 3

Total=18

Therefore total 18 task force can be formed.

7 0
4 years ago
Find the slope for the line that passes through (1,-10) that has slope - 9/5
soldi70 [24.7K]

Answer:

y=-9/5x-41/5

Step-by-step explanation:

y-y1=m(x-x1)

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y+10=-9/5(x-1)

y=-9/5x+9/5-10

y=-9/5x+9/5-50/5

y=-9/5x-41/5

6 0
3 years ago
Select the two binomials that are factors of this trinomial.
Yuki888 [10]

Answer:

b and d

Step-by-step explanation:

Sum = 4

Product = -32

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6 0
3 years ago
Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Otrada [13]

I guess the "5" is supposed to represent the integral sign?

I=\displaystyle\int_1^4\ln t\,\mathrm dt

With n=10 subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

\ell_i=1+\dfrac{3(i-1)}{10}

and right endpoints are given by

r_i=1+\dfrac{3i}{10}

where 1\le i\le10.

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, \dfrac{4-1}{10}=\dfrac3{10}, and "bases" equal to the values of \ln t at both endpoints of each subinterval. The area of the trapezoid over the i-th subinterval is

\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)

Then the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of \ln t at the average of the subinterval's endpoints, \dfrac{\ell_i+r_i}2. The area of the rectangle over the i-th subinterval is then

\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}

so the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}

c. For Simpson's rule, we find a quadratic interpolation of \ln t over each subinterval given by

P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}

where m_i is the midpoint of the i-th subinterval,

m_i=\dfrac{\ell_i+r_i}2

Then the integral I is equal to the sum of the integrals of each interpolation over the corresponding i-th subinterval.

I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt

It's easy to show that

\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)

so that the value of the overall integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}

4 0
4 years ago
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