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3 years ago

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You need to get to class, 200 meters away, and you can only walk in the hallways at about 1.5 m/s. (if you run any faster, you’l

l be caught for running). How much time will it take to get to your class?

2 answers:

RUDIKE [14]3 years ago

6 0

Answer:

Time = 133.3 secs

Step-by-step explanation:

<u><em>Speed = Distance / Time</em></u>

<em>Rearranging the formula</em>

Time = Distance / Speed

Where D = 200 m , S = 1.5 m/s

Time = 200 / 1.5

Time = 133.3 secs

meriva3 years ago

6 0

Answer:

133.3 seconds

Step-by-step explanation:

200/1.5 = 133.3

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Xavier has $23.50 in his savings account. He deposits $35 every week. His mother also deposits $20 into the account every time

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Answer:

Part A: a coefficient is either 20 or 35, both work. A variable is c or w, again either work. The constant is 23.50.

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Step-by-step explanation:

A coefficient goes in front of the variable (which is the letter, usually shown as x). The constant is a number with no variable attached.

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Lines e and f are parallel. The mAngle9 = 80° and mAngle5 = 55°. Parallel lines e and f are cut by transversal c and d. All angl

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Answer:

$(A)m\angle 2=125^\circ\\(C)m\angle 8=55^\circ\\(E)m\angle 14=100^\circ$

Step-by-step explanation:

The diagram of the problem is drawn and attached.

Given that:

$m\angle 9 =80^\circ\\m\angle 5 =55^\circ$

$m\angle 5+ m\angle 6=180^\circ\\55^\circ+ m\angle 6=180^\circ\\m\angle 6=180^\circ-55^\circ\\m\angle 6=125^\circ\\$Now:\\m\angle 6 = m\angle 2 $(Corresponging angles)\\Therefore m\angle 2=125^\circ\\$

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Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

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