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3 years ago

Which graph represents points on the polar curve r = 2 + 5sin(θ)?

Mathematics

2 answers:

solong [7]3 years ago

4 0

Answer:

Graph 2

Step-by-step explanation:

See image attached. Hope this helps.

VMariaS [17]3 years ago

3 0

Answer:

Step-by-step explanation:

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Let f(x)=x2 and g(x)=x-3. Evaluate (f•g)(0)

Galina-37 [17]

Answer:The answe would be 5 cause the 0 doesn't count

Step-by-step explanation:

3 0

2 years ago

Jonas wrote 9+8=17, what number sentence shows th commutative property of addition

Valentin [98]

The number sentence would be 8+9=17 because commutative property is when you just switch around the numbers but get the same answer.

7 0

3 years ago

Will mark Brainlest please answer. find the value of a,b. <br>,p,q from the equal order pairs

NARA [144]

Step-by-step explanation:

<h3>Question-1:</h3>

by order pair we obtain:

$\displaystyle \begin{cases} \displaystyle 3p = 2p - 1 \dots \dots i\\2q - p = 1 \dots \dots ii\end{cases}$

cancel 2p from the i equation to get a certain value of p:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q - p = 1 \end{cases}$

now substitute the value of p to the second equation:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q - ( - 1) = 1 \end{cases}$

simplify parentheses:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q + 1= 1 \end{cases}$

cancel 1 from both sides:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q = 0\end{cases}$

divide both sides by 2:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\q = 0\end{cases}$

<h3>question-2:</h3>

by order pair we obtain:

$\displaystyle \begin{cases} \displaystyle 2x - y= 3 \dots \dots i\\3y= x + y \dots \dots ii\end{cases}$

cancel out y from the second equation:

$\displaystyle \begin{cases} \displaystyle 2x - y= 3 \dots \dots i\\ x = 2y \dots \dots ii\end{cases}$

substitute the value of x to the first equation:

$\displaystyle \begin{cases} \displaystyle 2.2y-y= 3 \\ x = 2y \end{cases}$

simplify:

$\displaystyle \begin{cases} \displaystyle 3y= 3 \\ x = 2y \end{cases}$

divide both sides by 3:

$\displaystyle \begin{cases} \displaystyle y= 1 \\ x = 2y \end{cases}$

substitute the value of y to the second equation which yields:

$\displaystyle \begin{cases} \displaystyle y= 1 \\ x = 2 \end{cases}$

<h3>Question-3:</h3>

by order pair we obtain;

$\displaystyle \begin{cases} \displaystyle 2p + q = 2 \dots \dots i\\3q + 2p = 3 \dots \dots ii\end{cases}$

rearrange:

$\displaystyle \begin{cases} \displaystyle 2p + q = 2 \\2p + 3q= 3 \end{cases}$

subtract and simplify

$\displaystyle \begin{array}{ccc} \displaystyle 2p + q = 2 \\2p + 3q= 3 \\ \hline - 2q = - 1 \\ q = \dfrac{1}{2} \end{array}$

substitute the value of q to the first equation:

$\displaystyle 2.p+ \frac{1}{2} = 2$

make q the subject of the equation:

$\displaystyle p = \frac{3}{4}$

hence,

$\displaystyle q = \frac{1}{2} \\ p = \frac{3}{4}$

3 0

3 years ago

Read 2 more answers

Find the perimeter of the shape below:

nata0808 [166]

C is the answer i think

4 0

3 years ago

Read 2 more answers

I need to know how to do the whole thing and understand it.

patriot [66]

We are given the data on the number of candies handed by neighborhood A and neighborhood B.

Let us first find the mean and variance of each neighborhood.

Mean:

$\bar{x}_A=\frac{\sum x}{N_1}=\frac{12}{6}=2$ $\bar{x}_B=\frac{\sum x}{N_2}=\frac{20}{6}=3.33$

Variance:

$s_A^2=\frac{\sum x^2}{N_1}-\bar{x}_A^2=\frac{28}{6}-2^2=0.667$ $s_B^2=\frac{\sum x^2}{N_2}-\bar{x}_B^2=\frac{80}{6}-3.33^2=2.244$

A. Null hypothesis:

The null hypothesis is that there is no difference in the mean number of candies handed out by neighborhoods A and B.

$H_0:\;\mu_A=\mu_B$

Research hypothesis:

The research hypothesis is that the mean number of candies handed out by neighborhood A is more than neighborhood B.

$H_a:\;\mu_A>\mu_B$

Test statistic (t):

The test statistic of a two-sample t-test is given by

$t=\frac{\bar{x}_A-\bar{x}_B}{s_p}$

Where sp is the pooled standard deviation given by

$\begin{gathered} s_p=\sqrt{\frac{N_1s_1^2+N_2s_2^2}{N_1+N_2-2}(\frac{N_1+N_2}{N_1\cdot N_2}}) \\ s_p=\sqrt{\frac{6\cdot0.667+6\cdot2.244}{6+6-2}(\frac{6+6}{6\cdot6})} \\ s_p=0.763 \end{gathered}$ $t=\frac{2-3.33}{0.763}=-1.74$

So, the test statistic is -1.74

Critical t:

Degree of freedom = N1 + N2 - 2 = 6+6-2 = 10

Level of significance = 0.05

The right-tailed critical value for α = 0.05 and df = 10 is found to be 1.81

Critical t = 1.81

We will reject the null hypothesis because the calculated t-value is less than the critical value.

Interpretation:

This means that we do not have enough evidence to conclude that neighborhood A gives out more candies than neighborhood B.

6 0

1 year ago