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vazorg [7]
3 years ago
7

For rhombus LMNO, m∠LON = 102° and NP = 5 units. Use the diagram of rhombus LMNO to find the missing measures. The measure of ∠L

PM is °. The measure of ∠PMN is °. The length of LN is units.
Mathematics
2 answers:
aivan3 [116]3 years ago
8 0
LPM = 90

PMN = 51

LN = 10
Phantasy [73]3 years ago
8 0

Answer: angle LPM = 90, angle PMN = 51, side LN = 10

Explanation: Since, In case of rhombus, Diagonals bisect each other with the angle of 90, and corresponding angles are equal.

Here, In rhombus LMNO

\angle LON=\angle PMN = 102° (because both are corresponding angles)

⇒ \angle PMN =\angle LMN/2=102°/2

⇒\angle LPM =51°

Now, \angle LPM=90°

and, LN=LP+PN=5+5=10 unit.



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viktelen [127]

Answer:

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Step-by-step explanation:

Solve the following system:

{5 x + 3 y = 10 | (equation 1)

x = y - 6 | (equation 2)

Express the system in standard form:

{5 x + 3 y = 10 | (equation 1)

x - y = -6 | (equation 2)

Subtract 1/5 × (equation 1) from equation 2:

{5 x + 3 y = 10 | (equation 1)

0 x - (8 y)/5 = -8 | (equation 2)

Multiply equation 2 by -5/8:

{5 x + 3 y = 10 | (equation 1)

0 x+y = 5 | (equation 2)

Subtract 3 × (equation 2) from equation 1:

{5 x+0 y = -5 | (equation 1)

0 x+y = 5 | (equation 2)

Divide equation 1 by 5:

{x+0 y = -1 | (equation 1)

0 x+y = 5 | (equation 2)

Collect results:

Answer: {x = -1 , y = 5

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3 years ago
What are the minimum, first quartile, median, third quartile, and maximum of the data set? PLEASE Show your work.
Schach [20]
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First put your numbers in order from least to greatest. This shows your minimum is 54 and maximum is 72.
In the middle of the data set is 61.

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To find the first qaurtile only use the numbers before rhe median (54 55 59 61) and find the median of that. Since 55 and 59 are both in the middle add them both and divide by two to find the middle lf those two numbers. You should get 57. So 57 is the first quartile.

Do the same steps with the numbers after the meadian for your third quartile. You should get 69 as your answer for the third quartile.
7 0
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Step-by-step explanation:

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Solve the equation 2^(2x+1)+8=17×2^x.
myrzilka [38]
Oooh, looks fun

ok
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we might want to know some properties
(x^m)(x^n)=x^{m+n}
(a^b)^c=a^{bc}
if a^m=a^n where a=a, then assume m=n

2^{2x+1}=(2^{2x})(2^1)
so
(2^{2x})(2)+8=17(2^x)
(2^{2x})(2)+2^3=17(2^x)
(2^x)^2(2)+2^3=17(2^x)
minus 17*2^x both sides
(2^x)^2(2)+2^3-17(2^x)=0
use u subsitution, u=2ˣ
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or
2u^2-17u+8=0
ac method
2 times 8=16
what 2 number multiply to get -17 and add to get 16
-16 and -1
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(u-8)(2u-1)+0
u-8=0
u=8

2u-1=0
2u=1
u=1/2

now
u=2ˣ

8=2ˣ
2^3=2^x
3=x

and
\frac{1}{2}=2^x
2^{-1}=2^x
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2 years ago
Janet is training for a marathon. She decides to track her time per mile, in minutes, for 8 weeks. Which equation best fits the
Anna007 [38]

Answer:

y=-0.18 x+10.06

Step-by-step explanation:

The nine coordinates are (0,10), (1,9.9),  (2,9.6),(3,9.5), (4,9.4), (5,9.2), (6,9), (7,8.7), (8,8.6)

To find the line that best fit the data is to substitute the values in this equation y=a x+b where a=\frac{\left(N * \sum x y\right)-\left(\sum x * \sum y\right)}{\left(N * \sum x^{2}\right)-\left(\sum x * \sum x\right)} and b=\frac{\left(\sum x^{2} * \Sigma y\right)-\left(\sum x * \Sigma x y\right)}{\left(N * \Sigma x^{2}\right)-\left(\sum x * \Sigma x\right)}

Using the values from the table attached below, we can substitute and find the values of a and b

Substituting the values of a, we get,

\begin{aligned}a &=\frac{\left(N * \sum x y\right)-\left(\sum x * \Sigma y\right)}{\left(N * \sum x^{2}\right)-\left(\sum x * \sum x\right)} \\&=\frac{(9 * 324.9)-(36 * 83.9)}{(9 * 204)-(36 * 36)} \\&=-0.18\end{aligned}

Similarly, substituting the values of b, we get,

\begin{aligned}b &=\frac{\left(\sum x^{2} * \Sigma y\right)-\left(\sum x * \sum x y\right)}{\left(N * \sum x^{2}\right)-\left(\sum x * \sum x\right)} \\&=\frac{(204 * 83.9)-(36 * 315)}{(9 * 204)-(36 * 36)} \\&=10.06\end{aligned}

Thus, substituting the values of a and b in the formula, we get,

\begin{aligned}y &=a x+b \\&=-0.18 x+10.06\end{aligned}

8 0
3 years ago
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