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3 years ago

PLEASE HELP!!!! IMAGE ATTACHED FIND THE MEASURE OF ANGLE 3

Mathematics

1 answer:

Aleks04 [339]3 years ago

3 0

M< 3 = 180 - 60 - 90
m< 3 = 30

answer
30

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904.78cm^3

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What is the yintercept of f(x) =(1/2)^x<br> ?

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3 years ago

Rory earned an 84% on his test. He answered 21 questions correctly. How many total questions were on the test? Round to the near

Free_Kalibri [48]

Answer:

25 questions

Step-by-step explanation:

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3 years ago

the performance score of 10 adults is recorded, and the results are 83 87 90 92 93 100 104 111 115 121 find the standard deviati

luda_lava [24]

Answer:

The standard deviation of the data set is $\sigma = 12.7906$ .

Step-by-step explanation:

The Standard Deviation is a measure of how spread out numbers are. Its symbol is σ (the greek letter sigma)

To find the standard deviation of the following data set

$\begin{array}{cccccccc}83&87&90&92&93&100&104&111\\115&121&&&&&&\end{array}$

we use the following formula

$\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{X}\right)^2 }}{n-1} }$

Step 1: Find the mean $\left( \overline{X} \right)$ .

The mean of a data set is the sum of the terms divided by the total number of terms. Using math notation we have:

$Mean = \frac{Sum ~ of ~ terms}{Number ~ of ~ terms}$

$Mean = \frac{Sum ~ of ~ terms}{Number ~ of ~ terms}=\frac{83+87+90+92+93+100+104+111+115+121}{10} \\\\Mean = \frac{996}{10} =\frac{498}{5}=99.6$

Step 2: Create the below table.

Step 3: Find the sum of numbers in the last column to get.

$\sum{\left(x_i - \overline{X}\right)^2} = 1472.4$

Step 4: Calculate σ using the above formula.

$\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{X}\right)^2 }}{n-1} } = \sqrt{ \frac{ 1472.4 }{ 10 - 1} } \approx 12.7906$

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3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago