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3 years ago

Few math questions TIME IS RUNNING OUT!! PLEASE HELP!!!!

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

3 0

1.

$|5+6x|\ \textless \ 13\\\\-13\ \textless \ 5+6x\ \textless \ 13$

so

$\begin{cases}5+6x\ \textless \ 13\\5+6x\ \textgreater \ -13\end{cases}$

Answers B and D.

2.

$|x+9|\geq11\\\\x+9\ \textgreater \ 11\qquad\vee\qquad x+9\leq-11\\\\
\boxed{x\geq2\qquad\vee\qquad x\leq-20}$

Answer D.

3.

The same as point 2. Answer D.

4.

Answer C.

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3 years ago

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3 years ago

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Which of the following best represents the highest potential for nonresponse bias in a sampling strategy? Describe why this opti

natta225 [31]

Answer:

c. Asking people leaving a local election to take part in an exit poll

Step-by-step explanation:

Asking people leaving a local election to take part in an exit poll best represents the highest potential for nonresponse bias in a sampling strategy because of the importance of the local election compared to the exit polls.

It is worthy of note that nonresponse bias occurs when some respondents included in the sample do not respond to the survey. The major difference here is that the error comes from an absence of respondents not the collection of erroneous data. ...

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3 years ago

A 140 kg camera is suspended by two wires over a 40 metre wide football field to get shots of the action from above. At one poin

Katen [24]

Answer:

Please red the answer below

Step-by-step explanation:

In order to determine the length of each cable you use the Newton second law for each component of the forces involved in the situation.

For the x component you have:

$T_1cos\theta_1-T_2cos\theta_2=0$ (1)

T1: tension of the first cable = 1500N

T2: tension of the second cable = 800N

θ1: angle between the horizontal and the first cable

θ2: angle between the horizontal and the first cable

For the y component you have:

$T_1sin\theta_1+T_2sin\theta_2-W=0$ (2)

W: weight of the camera = Mg = (140kg)(9.8m/s^2) = 1372N

You can squared both equations (1) and (2) and the sum the two equations:

$T_1^2cos^2\theta_1=T_2^2cos^2\theta_2\\\\T_1^2sin^2\theta_1=T_2^2sin^2\theta_2-2WT_2sin\theta_2+W^2$

Then, you sum the equations:

$T_1^2(cos^2\theta_1+sin^2\theta_1)=T_2^2(sin^2\theta_2+cos^2\theta_2)-2Wsin\theta_2+W^2$ (3)

Next, you use the following identity:

$sin^2\theta+cos^2\theta=1$

and you obtain in the equation (3):

$T_1^2=T_2^2-2WT_2sin\theta_2+W^2\\\\sin\theta_2=\frac{T_2^2-T_1^2+W^2}{2WT_2}=\frac{(800N)^2-(1500)^2+(1372N)^2}{2(800N)(1372N)}=0.066\\\\\theta_2=sin^{-1}0.066=27.23\°$

With this values you can calculate the value of the another angle, by using the equation (1):

$\theta_1=cos^{-1}(\frac{T_2cos(27.23\°)}{T_1})=cos^{-1}(\frac{(800N)(cos27.23\°)}{1500N})\\\\\theta_1=61.69\°$

Now, you can calculate the length of each cable by using the information about the width of the football field. You use the following trigonometric relation:

$l_1cos\theta_1=40-d\\\\l_2cos\theta_2=d\\\\$

d: distance to the right side of the field

By using the cosine law you can fins a system of equation and then you can calculate the values of l1 and l2.

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2 years ago

Need some help with the answer to the question

kicyunya [14]

Answer:

(- 4, 27 )

Step-by-step explanation:

Equate the right sides of both equations, that is

x² - 2x + 3 = - 2x + 19 ← subtract - 2x + 19 from both sides

x² - 16 = 0 ← in standard form

(x - 4)(x + 4) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 4 = 0 ⇒ x = 4

x + 4 = 0 ⇒ x = - 4

Substitute these values into f(x) = - 2x + 19

f(4) = - 2(4) + 19 = - 8 + 19 = 11 ⇒ (4, 11 )

f(- 4) = - 2(- 4) + 19 = 8 + 19 = 27 ⇒ (- 4, 27 )

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2 years ago