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3 years ago

Can someone please help me with this geometry question.

Mathematics

2 answers:

Scrat [10]3 years ago

8 0

W is 63 . X is 86 . Y is 38 . Z is 25

I hope this helps!

galben [10]3 years ago

6 0

To find w, you subtract 117 from 180. After that you get w = 63. Now you add 63 and 31 to get 94. You then subtract 94 from 180 to get x = 86. Now you add 86 and 56 to get 142. You then subtract 142 from 180 to get y = 38. Now you add 38 and 117 to get 155. Again you subtract 155 from 180 to get z = 25

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A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

4 0

3 years ago

Sketch the graph by hand using asymptotes and intercepts, but not derivatives. Then use your sketch as a guide to producing grap

Arisa [49]

The local minima of $f\left(x\right)=\ \frac{\left(2x+3\right)^2\left(x\ -2\right)^5}{x^3\left(x-5\right)^2}$ are (x, f(x)) = (-1.5, 0) and (7.980, 609.174)

<h3>How to determine the local minima?</h3>

The function is given as:

$f\left(x\right)=\ \frac{\left(2x+3\right)^2\left(x\ -2\right)^5}{x^3\left(x-5\right)^2}$

See attachment for the graph of the function f(x)

From the attached graph, we have the following minima:

Minimum = (-1.5, 0)

Minimum = (7.980, 609.174)

The above means that, the local minima are

(x, f(x)) = (-1.5, 0) and (7.980, 609.174)

Read more about graphs at:

brainly.com/question/20394217

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2 years ago

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Eddi Din [679]

Answer:

Step-by-step explanation:

It says Renita takes minutes to ride half a mile.

She leaves 30 minutes before school.

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Since she leaves 30minutes before school. She will reach 5minutes early before school.

8 0

3 years ago

You are considering developing a regression equation relating a dependent variable to two independent variables. One of the vari

lys-0071 [83]

Answer:

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We need 1 less dummy variable than outcomes for the categorical variable.

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Answer: 1

Note: If our categorical variable had 5 outcomes, we would need 5-1 = 4 dummy variables.

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Note: Depending your your professor, this can be written one of the following ways...

yhat = a + b1 x1 + bx2

yhat = b0 + b1 x1 + b2 x2

yhat = beta0hat + beta1hat x1 + betahat2 x2

where x1 is the ratio variable

and x2 is the dummy variable

c. Interpret the meaning of the coefficients in the regression equation.

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3 years ago

What number should be added to both sides of the equation to complete the square x^2+12x=11

maksim [4K]

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3 years ago