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fenix001 [56]
3 years ago
5

Please help, this is geometry

Mathematics
1 answer:
maw [93]3 years ago
3 0

Answer:

4  2/9π   or   4.22 (repeating)π

Step-by-step explanation:

<em>I changed my mind. I'll attempt it.</em>

The formula for finding the arc length: 2\pi (r)(\frac{theta}{360} )

2π(4)(190/360) = 4 2/9π

82% certain the answer you're looking for is 4  2/9π (<em>pi symbol because it asks for it in terms of pi)</em>

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What is the solution to the inequality 7&lt;=-3x-4
kipiarov [429]

Answer:

The nearest option is -3 (negative 3)

So, Option B is correct.

Step-by-step explanation:

We need to solve the inequality 7 \leq  -3x-4

Solving:

Step 1: Switching sides of inequality and reversing the inequality

-3x-4\geq 7

Step 2: Adding 4 on both sides

-3x-4+4\geq 7+4\\-3x\geq 11

Step 3: Dividing both sides by -3 and reversing the inequality

\frac{-3x}{-3}\leq \frac{11}{-3}  \\x\leq -3.6\\

The nearest option is -3 (negative 3)

So, Option B is correct.

4 0
3 years ago
The daily revenues of a cafe near the university are approximately normally distributed. The owner recently collected a random s
lbvjy [14]

Answer:

The sample size to obtain the desired margin of error is 160.

Step-by-step explanation:

The Margin of Error is given as

MOE=z_{crit}\times\dfrac{\sigma}{\sqrt{n}}

Rearranging this equation in terms of n gives

n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2

Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as

n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n

As n is given as 40 so the new sample size is given as

n_2=4n\\n_2=4*40\\n_2=160

So the sample size to obtain the desired margin of error is 160.

4 0
3 years ago
Find the surface area of the cylinder
Stels [109]

Answer:

Step-by-step explanation:

مرحبًا ، أود أن أحاول مساعدتك في عملك المدرسي. إذا كنت تستطيع أن تشرح أكثر فأنا أقدر ذلك.

3 0
3 years ago
What is 6,4 rounded to the nearest whole number
Aleks04 [339]
6.4 rounded to the nearest whole number will be 6.

if the decimal was 5 or larger, it would’ve been 7.
6 0
2 years ago
Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is
ohaa [14]

The vectors in S form a basis of P_2 if they are mutually linearly independent and span P_2.

To check for independence, we can compute the Wronskian determinant:

\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0

The determinant is non-zero, so the vectors are indeed independent.

To check if they span P_2, you need to show that any vector in P_2 can be expressed as a linear combination of the vectors in S. We can write an arbitrary vector in P_2 as

p=ax^2+bx+c

Then we need to show that there is always some choice of scalars k_1,k_2,k_3 such that

k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p

This is equivalent to solving

(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c

or the system (in matrix form)

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

This has a solution if the coefficient matrix on the left is invertible. It is, because

\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}

Then

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}

\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}

A solution exists for any choice of a,b,c, so the vectors in S indeed span P_2.

The vectors in S are independent and span P_2, so S forms a basis of P_2.

5 0
3 years ago
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